Veri yap\u0131lar\u0131 ve algoritmalar d\u00fcnyas\u0131nda y\u0131\u011f\u0131n (stack), belki de en temel ama ayn\u0131 zamanda en g\u00fc\u00e7l\u00fc ara\u00e7lardan biridir. Peki, bilgisayarlar\u0131n g\u00fcnl\u00fck hayatta kar\u015f\u0131la\u015ft\u0131\u011f\u0131m\u0131z karma\u015f\u0131k g\u00f6revleri nas\u0131l y\u00f6netti\u011fini hi\u00e7 merak ettiniz mi? Taray\u0131c\u0131lar\u0131n\u0131z\u0131n geri tu\u015fu nas\u0131l \u00e7al\u0131\u015f\u0131r, metin d\u00fczenleyicilerdeki “geri al” (undo) fonksiyonu hangi prensibe dayan\u0131r ya da bir program\u0131n fonksiyon \u00e7a\u011fr\u0131lar\u0131 bellekte nas\u0131l organize edilir? T\u00fcm bu sorular\u0131n cevab\u0131, basit ama etkili bir veri yap\u0131s\u0131 olan y\u0131\u011f\u0131nda sakl\u0131d\u0131r. Bu makalede, y\u0131\u011f\u0131n\u0131n teorik temellerinden ba\u015flayarak, ger\u00e7ek d\u00fcnya senaryolar\u0131ndaki uygulamalar\u0131na ve LeetCode gibi platformlardaki problem \u00e7\u00f6zme pratiklerine kadar derinlemesine bir yolculu\u011fa \u00e7\u0131kaca\u011f\u0131z. Kod \u00f6rnekleri ve ad\u0131m ad\u0131m a\u00e7\u0131klamalarla, y\u0131\u011f\u0131n\u0131 s\u0131f\u0131rdan \u00f6\u011frenip, algoritmik d\u00fc\u015f\u00fcnme becerilerinizi geli\u015ftireceksiniz.<\/p>\n
Bilgisayar bilimlerinde bir y\u0131\u011f\u0131n, belirli bir prensibe g\u00f6re \u00e7al\u0131\u015fan, soyut bir veri yap\u0131s\u0131d\u0131r. T\u0131pk\u0131 \u00fcst \u00fcste dizilmi\u015f tabaklar veya kitaplar gibi d\u00fc\u015f\u00fcnebilirsiniz. Yeni bir eleman ekledi\u011finizde her zaman en \u00fcste gelirken, bir eleman\u0131 \u00e7\u0131karmak istedi\u011finizde ise sadece en \u00fcstteki eleman\u0131 alabilirsiniz. Bu karakteristik \u00f6zellik, y\u0131\u011f\u0131n\u0131 di\u011fer veri yap\u0131lar\u0131ndan ay\u0131ran en temel ilkedir: “Son Giren \u0130lk \u00c7\u0131kar” (Last-In, First-Out – LIFO). Yani, y\u0131\u011f\u0131na en son eklenen eleman, y\u0131\u011f\u0131ndan ilk \u00e7\u0131kan eleman olacakt\u0131r.<\/p>\n
Y\u0131\u011f\u0131n\u0131n bu LIFO prensibi, bir\u00e7ok ger\u00e7ek d\u00fcnya ve programlama senaryosunda kritik bir rol oynar. \u00d6rne\u011fin, bir web taray\u0131c\u0131s\u0131n\u0131n ge\u00e7mi\u015fini d\u00fc\u015f\u00fcn\u00fcn. Bir sayfay\u0131 ziyaret etti\u011finizde, bu sayfa ge\u00e7mi\u015f y\u0131\u011f\u0131n\u0131na eklenir. “Geri” tu\u015funa bast\u0131\u011f\u0131n\u0131zda ise, en son ziyaret etti\u011finiz sayfa (yani y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki eleman) y\u00fcklenir ve y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131l\u0131r. Bu basit mekanizma, karma\u015f\u0131k kullan\u0131c\u0131 deneyimlerinin temelini olu\u015fturur. Benzer \u015fekilde, metin d\u00fczenleyicilerdeki “geri al” (undo) i\u015flevi de y\u0131\u011f\u0131n mant\u0131\u011f\u0131yla \u00e7al\u0131\u015f\u0131r. Her yapt\u0131\u011f\u0131n\u0131z i\u015flem bir y\u0131\u011f\u0131na eklenir ve “geri al” dedi\u011finizde en son i\u015flemi geri al\u0131rs\u0131n\u0131z.<\/p>\n
Peki, y\u0131\u011f\u0131n neden bu kadar \u00f6nemlidir? \u00d6ncelikle, karma\u015f\u0131kl\u0131\u011f\u0131 y\u00f6netmeye yard\u0131mc\u0131 olur. Bir\u00e7ok algoritmik problem, y\u0131\u011f\u0131n\u0131n LIFO yap\u0131s\u0131 sayesinde daha zarif ve verimli bir \u015fekilde \u00e7\u00f6z\u00fclebilir. Fonksiyon \u00e7a\u011fr\u0131lar\u0131n\u0131n y\u00f6netimi gibi temel bilgisayar operasyonlar\u0131 bile y\u0131\u011f\u0131n kullan\u0131larak ger\u00e7ekle\u015ftirilir. Bir program bir fonksiyonu \u00e7a\u011f\u0131rd\u0131\u011f\u0131nda, bu fonksiyonun bilgileri (parametreler, yerel de\u011fi\u015fkenler vb.) bir “\u00e7a\u011fr\u0131 y\u0131\u011f\u0131n\u0131na” (call stack) itilir. Fonksiyon \u00e7al\u0131\u015fmas\u0131n\u0131 bitirdi\u011finde ise, bu bilgiler y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131l\u0131r ve kontrol \u00e7a\u011f\u0131ran fonksiyona geri d\u00f6ner. Bu mekanizma, programlar\u0131n d\u00fczenli ve \u00f6ng\u00f6r\u00fclebilir bir \u015fekilde \u00e7al\u0131\u015fmas\u0131n\u0131 sa\u011flar.<\/p>\n
\u00d6zellikle algoritmalar ve veri yap\u0131lar\u0131 d\u00fcnyas\u0131na yeni ad\u0131m atanlar i\u00e7in y\u0131\u011f\u0131n, \u00f6\u011frenilmesi gereken ilk ve en \u00f6nemli yap\u0131lardan biridir. Anlamas\u0131 kolay, uygulamas\u0131 nispeten basit olmakla birlikte, karma\u015f\u0131k problemlerin \u00e7\u00f6z\u00fcm\u00fcnde esnek ve g\u00fc\u00e7l\u00fc bir ara\u00e7 sunar. Bu temel prensibi kavrad\u0131\u011f\u0131m\u0131zda, dinamik programlama, graf algoritmalar\u0131 ve hatta yapay zeka gibi daha ileri konularda kar\u015f\u0131m\u0131za \u00e7\u0131kacak olan bir\u00e7ok yap\u0131n\u0131n mant\u0131\u011f\u0131n\u0131 anlamak \u00e7ok daha kolay hale gelecektir. Bu nedenle, y\u0131\u011f\u0131n sadece bir veri yap\u0131s\u0131 olmaktan \u00f6te, programlaman\u0131n temel ta\u015flar\u0131ndan biridir ve yaz\u0131l\u0131m geli\u015ftirme yolculu\u011funuzda size s\u0131k\u00e7a e\u015flik edecektir.<\/p>\n
Her veri yap\u0131s\u0131n\u0131n kendine \u00f6zg\u00fc bir \u00e7al\u0131\u015fma prensibi ve bir dizi temel i\u015flemi vard\u0131r. Y\u0131\u011f\u0131n i\u00e7in bu prensip “Son Giren \u0130lk \u00c7\u0131kar” (LIFO) iken, temel i\u015flemler de y\u0131\u011f\u0131n\u0131n do\u011fas\u0131na uygun olarak tasarlanm\u0131\u015ft\u0131r. Bu b\u00f6l\u00fcmde, y\u0131\u011f\u0131n\u0131n kalbinde yer alan bu i\u015flemleri detayl\u0131 bir \u015fekilde inceleyecek ve y\u0131\u011f\u0131n\u0131n nas\u0131l \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131 daha iyi kavrayaca\u011f\u0131z.<\/p>\n
Y\u0131\u011f\u0131n\u0131n iki ana ve en s\u0131k kullan\u0131lan i\u015flemi vard\u0131r: Bu iki i\u015flemin yan\u0131 s\u0131ra, y\u0131\u011f\u0131n veri yap\u0131s\u0131n\u0131n etkinli\u011fini ve kullan\u0131\u015fl\u0131l\u0131\u011f\u0131n\u0131 art\u0131ran ba\u015fka yard\u0131mc\u0131 i\u015flemler de bulunmaktad\u0131r:<\/p>\n Y\u0131\u011f\u0131n soyut bir veri yap\u0131s\u0131 oldu\u011fu i\u00e7in, bilgisayar belle\u011finde farkl\u0131 \u015fekillerde uygulanabilir. En yayg\u0131n iki uygulama y\u00f6ntemi diziler (veya dinamik diziler) ve ba\u011fl\u0131 listelerdir.<\/p>\n Dizi Tabanl\u0131 Uygulama:<\/strong><\/p>\n Y\u0131\u011f\u0131n\u0131 bir dizi (array) kullanarak uygulamak olduk\u00e7a basittir. Dizinin sonunu y\u0131\u011f\u0131n\u0131n “\u00fcst” k\u0131sm\u0131 olarak kabul ederiz.<\/p>\n Dizi tabanl\u0131 y\u0131\u011f\u0131nlar, elemanlara do\u011frudan indeksle eri\u015fim sa\u011flad\u0131\u011f\u0131 i\u00e7in genellikle h\u0131zl\u0131d\u0131r. Ancak, sabit boyutlu diziler kullan\u0131ld\u0131\u011f\u0131nda kapasite a\u015f\u0131m\u0131 (overflow) riski ta\u015f\u0131r. Dinamik diziler (bellek gerekti\u011finde kendini yeniden boyutland\u0131ran diziler) bu sorunu \u00e7\u00f6zer ancak ara s\u0131ra yeniden boyutland\u0131rma maliyeti olabilir.<\/p>\n Ba\u011fl\u0131 Liste Tabanl\u0131 Uygulama:<\/strong><\/p>\n Bir ba\u011fl\u0131 liste kullanarak y\u0131\u011f\u0131n uygulamak da pop\u00fcler bir y\u00f6ntemdir. Bu durumda, ba\u011fl\u0131 listenin ba\u015f\u0131 (head) y\u0131\u011f\u0131n\u0131n “\u00fcst” k\u0131sm\u0131n\u0131 temsil eder.<\/p>\n Ba\u011fl\u0131 liste tabanl\u0131 y\u0131\u011f\u0131nlar\u0131n en b\u00fcy\u00fck avantaj\u0131, dinamik boyutland\u0131rma yetene\u011fidir; yani, y\u0131\u011f\u0131n\u0131n boyutu bellek izin verdi\u011fi s\u00fcrece s\u0131n\u0131rs\u0131zca b\u00fcy\u00fcyebilir. Ancak, her eleman i\u00e7in ek bir bellek alan\u0131 (pointerlar i\u00e7in) gerektirir ve dizi tabanl\u0131 uygulamalara k\u0131yasla elemanlara eri\u015fim biraz daha dolayl\u0131 olabilir.<\/p>\n Her iki uygulama y\u00f6ntemi de LIFO prensibini korur ve \u00e7o\u011fu durumda O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131 sunar. Se\u00e7im, genellikle performans gereksinimlerine, bellek k\u0131s\u0131tlamalar\u0131na ve geli\u015ftirme kolayl\u0131\u011f\u0131na ba\u011fl\u0131d\u0131r. Python gibi dillerde, yerle\u015fik listeler zaten dinamik diziler oldu\u011fu i\u00e7in, y\u0131\u011f\u0131n uygulamak \u00e7ok kolay ve verimlidir.<\/p>\n Y\u0131\u011f\u0131n veri yap\u0131s\u0131, sadece teorik bir kavram olman\u0131n \u00f6tesinde, g\u00fcnl\u00fck kulland\u0131\u011f\u0131m\u0131z bir\u00e7ok yaz\u0131l\u0131m\u0131n ve sistemin temelinde yatan kritik bir bile\u015fendir. Soyut bir kavram gibi g\u00f6r\u00fcnse de, y\u0131\u011f\u0131n\u0131n LIFO prensibi, karma\u015f\u0131k s\u00fcre\u00e7leri d\u00fczenlemek ve y\u00f6netmek i\u00e7in inan\u0131lmaz derecede esneklik sunar. Gelin, y\u0131\u011f\u0131n\u0131n ger\u00e7ek d\u00fcnyada nas\u0131l kullan\u0131ld\u0131\u011f\u0131na dair baz\u0131 \u00e7arp\u0131c\u0131 \u00f6rnekleri ve bir vaka analizini birlikte inceleyelim.<\/p>\n Web Taray\u0131c\u0131lar\u0131 ve Gezinme Ge\u00e7mi\u015fi:<\/strong> Belki de y\u0131\u011f\u0131n\u0131n en bilinen ve en kolay anla\u015f\u0131l\u0131r pratik uygulamalar\u0131ndan biri web taray\u0131c\u0131lar\u0131m\u0131zd\u0131r. Herhangi bir web sayfas\u0131na gitti\u011finizde, bu sayfan\u0131n URL’si dahili bir y\u0131\u011f\u0131na ( Metin D\u00fczenleyicilerde “Geri Al” (Undo) ve “Yinele” (Redo):<\/strong> Bir metin d\u00fczenleyicide (Microsoft Word, Google Docs, Sublime Text vb.) yaz\u0131 yazarken, s\u00fcrekli olarak “geri al” veya “yinele” fonksiyonlar\u0131n\u0131 kullan\u0131r\u0131z. Bu i\u015flevler de y\u0131\u011f\u0131n prensibiyle \u00e7al\u0131\u015f\u0131r. Her yapt\u0131\u011f\u0131n\u0131z de\u011fi\u015fiklik (bir karakter yazmak, bir paragraf silmek, metin format\u0131n\u0131 de\u011fi\u015ftirmek) bir “i\u015flem y\u0131\u011f\u0131n\u0131na” (undo stack) kaydedilir. “Geri al” tu\u015funa bast\u0131\u011f\u0131n\u0131zda, en son yap\u0131lan i\u015flem bu y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131l\u0131r ve ilgili de\u011fi\u015fiklik geri al\u0131n\u0131r. Geri al\u0131nan i\u015flem ise genellikle ayr\u0131 bir “yinele y\u0131\u011f\u0131n\u0131na” (redo stack) eklenir. B\u00f6ylece, “yinele” tu\u015funa bast\u0131\u011f\u0131n\u0131zda, bu y\u0131\u011f\u0131ndan en son geri al\u0131nan i\u015flem \u00e7\u0131kar\u0131l\u0131r ve tekrar uygulan\u0131r. Bu iki y\u0131\u011f\u0131n, kullan\u0131c\u0131n\u0131n yapt\u0131\u011f\u0131 de\u011fi\u015fiklikleri esnek bir \u015fekilde y\u00f6netmesini sa\u011flar.<\/p>\n Bu mekanizma, basit gibi g\u00f6r\u00fcnse de, kullan\u0131c\u0131 deneyimi a\u00e7\u0131s\u0131ndan hayati \u00f6neme sahiptir. Kullan\u0131c\u0131ya hata yapma \u00f6zg\u00fcrl\u00fc\u011f\u00fc tan\u0131r ve karma\u015f\u0131k d\u00fczenleme i\u015flemlerinde g\u00fcvence sa\u011flar. D\u00fc\u015f\u00fcn\u00fcn ki, yanl\u0131\u015fl\u0131kla \u00f6nemli bir metni sildiniz ve geri alma \u00f6zelli\u011fi olmasayd\u0131. \u0130\u015fte tam da bu noktada y\u0131\u011f\u0131n\u0131n g\u00fcc\u00fc ve \u00f6nemi devreye giriyor.<\/p>\n Fonksiyon \u00c7a\u011fr\u0131lar\u0131 Y\u00f6netimi (Call Stack):<\/strong> Bilgisayar bilimindeki belki de en temel ve en kritik y\u0131\u011f\u0131n uygulamas\u0131, programlar\u0131n fonksiyon \u00e7a\u011fr\u0131lar\u0131n\u0131 y\u00f6netmek i\u00e7in kulland\u0131\u011f\u0131 “\u00e7a\u011fr\u0131 y\u0131\u011f\u0131n\u0131d\u0131r” (call stack). Bir program \u00e7al\u0131\u015f\u0131rken, bir fonksiyonu \u00e7a\u011f\u0131rd\u0131\u011f\u0131nda, bu fonksiyonun parametreleri, yerel de\u011fi\u015fkenleri ve d\u00f6n\u00fc\u015f adresi gibi bilgileri \u00e7a\u011fr\u0131 y\u0131\u011f\u0131n\u0131na ( Parantez E\u015fle\u015ftirme ve \u0130fade De\u011ferlendirme:<\/strong> Derleyiciler ve yorumlay\u0131c\u0131lar, kodunuzdaki parantezlerin ( G\u00f6r\u00fcld\u00fc\u011f\u00fc \u00fczere, y\u0131\u011f\u0131n sadece soyut bir veri yap\u0131s\u0131 olmaktan \u00e7ok daha fazlas\u0131d\u0131r. Modern yaz\u0131l\u0131m geli\u015ftirmenin, kullan\u0131c\u0131 deneyiminin ve hatta donan\u0131m d\u00fczeyindeki operasyonlar\u0131n ayr\u0131lmaz bir par\u00e7as\u0131d\u0131r. Bu pratik \u00f6rnekler, y\u0131\u011f\u0131n\u0131n algoritmik d\u00fc\u015f\u00fcnme becerilerimizi nas\u0131l zenginle\u015ftirdi\u011fini ve kar\u015f\u0131la\u015ft\u0131\u011f\u0131m\u0131z problemleri \u00e7\u00f6zmede ne kadar g\u00fc\u00e7l\u00fc bir ara\u00e7 olabilece\u011fini a\u00e7\u0131k\u00e7a ortaya koymaktad\u0131r.<\/p>\n Teorik bilgileri peki\u015ftirmenin en iyi yolu, \u00f6\u011frendiklerimizi prati\u011fe d\u00f6kmektir. Bu b\u00f6l\u00fcmde, Python programlama dilini kullanarak kendi y\u0131\u011f\u0131n (Stack) veri yap\u0131m\u0131z\u0131 ad\u0131m ad\u0131m nas\u0131l olu\u015fturaca\u011f\u0131m\u0131z\u0131 g\u00f6rece\u011fiz. Python’\u0131n listeleri, dinamik dizi \u00f6zellikleriyle y\u0131\u011f\u0131n implementasyonu i\u00e7in olduk\u00e7a uygundur ve bize kolayl\u0131k sa\u011flar. Kendi s\u0131n\u0131f\u0131m\u0131z\u0131 olu\u015fturarak y\u0131\u011f\u0131n\u0131n temel i\u015flemlerini (push, pop, peek, isEmpty, size) hayata ge\u00e7irece\u011fiz.<\/p>\n \u0130lk olarak, <\/code><\/p>\n Yukar\u0131daki kod blo\u011fu, \u015eimdi de bu <\/code><\/p>\n Bu \u00f6rnekte, Y\u0131\u011f\u0131n\u0131n teorik temellerini ve pratik uygulamas\u0131n\u0131 kavrad\u0131ktan sonra, s\u0131ra ger\u00e7ek d\u00fcnya problem \u00e7\u00f6zme becerilerinizi geli\u015ftirmeye geliyor. LeetCode, HackerRank gibi platformlar, algoritmalar\u0131 ve veri yap\u0131lar\u0131n\u0131 kullanarak problem \u00e7\u00f6zme yetene\u011finizi test etmek ve geli\u015ftirmek i\u00e7in m\u00fckemmel kaynaklard\u0131r. Y\u0131\u011f\u0131n, bu platformlarda kar\u015f\u0131n\u0131za \u00e7\u0131kacak bir\u00e7ok problemin anahtar \u00e7\u00f6z\u00fcm mekanizmas\u0131d\u0131r. Bu b\u00f6l\u00fcmde, y\u0131\u011f\u0131n ile \u00e7\u00f6z\u00fclebilecek tipik LeetCode problemlerine genel bir bak\u0131\u015f atacak ve pop\u00fcler bir \u00f6rne\u011fi detayland\u0131raca\u011f\u0131z.<\/p>\n Y\u0131\u011f\u0131nla \u00e7\u00f6z\u00fclen problemler genellikle \u015fu kategorilere ayr\u0131l\u0131r:<\/p>\n LeetCode'da s\u0131k\u00e7a kar\u015f\u0131la\u015f\u0131lan ve y\u0131\u011f\u0131n\u0131n g\u00fcc\u00fcn\u00fc g\u00f6steren klasik bir problem \"Ge\u00e7erli Parantezler\" (Valid Parentheses) problemidir. Problem \u015fu \u015fekildedir: Bir string S verildi\u011finde, bu stringin ge\u00e7erli parantezlerden olu\u015fup olu\u015fmad\u0131\u011f\u0131n\u0131 belirleyin. Ge\u00e7erli bir string \u015fu ko\u015fullar\u0131 sa\u011flamal\u0131d\u0131r:<\/p>\n \u00d6rnekler: \"()\" ge\u00e7erli, \"()[]{}\" ge\u00e7erli, \"(]\" ge\u00e7ersiz, \"([)]\" ge\u00e7ersiz.<\/p>\n Y\u0131\u011f\u0131n Yakla\u015f\u0131m\u0131:<\/strong><\/p>\n Bu problemi \u00e7\u00f6zmek i\u00e7in bir y\u0131\u011f\u0131n kullanabiliriz. Mant\u0131k olduk\u00e7a basittir:<\/p>\n <\/code><\/p>\n Bu kod par\u00e7as\u0131, \"Ge\u00e7erli Parantezler\" problemini y\u0131\u011f\u0131n kullanarak nas\u0131l \u00e7\u00f6zebilece\u011finizi ad\u0131m ad\u0131m g\u00f6stermektedir. Bu t\u00fcr problemler, y\u0131\u011f\u0131n\u0131n LIFO prensibini ve basitli\u011fini kullanarak karma\u015f\u0131k g\u00f6r\u00fcnen senaryolar\u0131 nas\u0131l \u00e7\u00f6zd\u00fc\u011f\u00fcm\u00fcz\u00fc anlamak i\u00e7in m\u00fckemmel \u00f6rneklerdir. LeetCode'da benzer bir\u00e7ok problem bulabilir, y\u0131\u011f\u0131n becerilerinizi daha da geli\u015ftirebilirsiniz. Unutmay\u0131n, pratik yapmak, algoritmalar\u0131 ve veri yap\u0131lar\u0131n\u0131 ustala\u015fman\u0131n anahtar\u0131d\u0131r.<\/p>\n Y\u0131\u011f\u0131n, temel bir veri yap\u0131s\u0131 olsa da, performans\u0131n\u0131 optimize etmek ve s\u0131n\u0131r durumlar\u0131 do\u011fru \u015fekilde ele almak, \u00f6zellikle b\u00fcy\u00fck \u00f6l\u00e7ekli uygulamalar veya rekabet\u00e7i programlama ortamlar\u0131 i\u00e7in hayati \u00f6nem ta\u015f\u0131r. Y\u0131\u011f\u0131n\u0131n basitli\u011finin ard\u0131nda, dikkat edilmesi gereken baz\u0131 ince detaylar ve ileri d\u00fczey ipu\u00e7lar\u0131 bulunur.<\/p>\n Bir y\u0131\u011f\u0131n\u0131n performans\u0131n\u0131 de\u011ferlendirirken, i\u015flemlerin zaman ve yer (bellek) karma\u015f\u0131kl\u0131\u011f\u0131n\u0131 anlamak \u00f6nemlidir. Y\u0131\u011f\u0131n\u0131n temel i\u015flemleri olan Ancak, bu O(1) durumu, y\u0131\u011f\u0131n\u0131n dinamik bir dizi (Python listeleri gibi) olarak uyguland\u0131\u011f\u0131 ve dizinin yeniden boyutland\u0131r\u0131lma ihtiyac\u0131 duymad\u0131\u011f\u0131 s\u00fcrece ge\u00e7erlidir. E\u011fer dizi tabanl\u0131 bir y\u0131\u011f\u0131n kullan\u0131yorsan\u0131z ve dizi kapasitesini a\u015farsan\u0131z, sistemin yeni, daha b\u00fcy\u00fck bir dizi olu\u015fturmas\u0131 ve mevcut elemanlar\u0131 bu yeni diziye kopyalamas\u0131 gerekebilir. Bu yeniden boyutland\u0131rma i\u015flemi, O(N) zaman karma\u015f\u0131kl\u0131\u011f\u0131na neden olabilir ve bu da i\u015flem s\u00fcresinde anl\u0131k bir yava\u015flamaya yol a\u00e7abilir. Ancak amortize edilmi\u015f (ortalama) zaman karma\u015f\u0131kl\u0131\u011f\u0131 yine de O(1) olarak kabul edilir, \u00e7\u00fcnk\u00fc bu t\u00fcr yeniden boyutland\u0131rmalar nadiren ger\u00e7ekle\u015fir.<\/p>\n Her veri yap\u0131s\u0131nda oldu\u011fu gibi, y\u0131\u011f\u0131nda da s\u0131n\u0131r durumlar (edge cases) mevcuttur ve bunlar\u0131n do\u011fru \u015fekilde ele al\u0131nmas\u0131 gerekmektedir. En yayg\u0131n s\u0131n\u0131r durumlar\u0131 \u015funlard\u0131r:<\/p>\n Baz\u0131 \u00f6zel problem t\u00fcrleri i\u00e7in standart y\u0131\u011f\u0131n yap\u0131s\u0131n\u0131 geli\u015ftirmek gerekebilir. \u00d6rne\u011fin, \"Min Stack\" (Minimum Y\u0131\u011f\u0131n) problemi, y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki eleman\u0131 O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131yla d\u00f6nd\u00fcrmenin yan\u0131 s\u0131ra, y\u0131\u011f\u0131ndaki en k\u00fc\u00e7\u00fck eleman\u0131 da O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131yla d\u00f6nd\u00fcrmesini ister. Bu, genellikle iki y\u0131\u011f\u0131n kullanarak \u00e7\u00f6z\u00fcl\u00fcr: bir ana y\u0131\u011f\u0131n ve sadece minimum de\u011ferleri tutan yard\u0131mc\u0131 bir y\u0131\u011f\u0131n. Bu t\u00fcr optimize edilmi\u015f tasar\u0131mlar, problemin gereksinimlerine g\u00f6re y\u0131\u011f\u0131n yap\u0131s\u0131n\u0131n esnekli\u011fini ve uyarlanabilirli\u011fini g\u00f6sterir.<\/p>\n Performans ve s\u0131n\u0131r durumlar\u0131 anlamak, sadece kodunuzun do\u011fru \u00e7al\u0131\u015fmas\u0131n\u0131 sa\u011flamakla kalmaz, ayn\u0131 zamanda daha sa\u011flam, g\u00fcvenilir ve verimli yaz\u0131l\u0131mlar geli\u015ftirmenize olanak tan\u0131r. Her zaman, belirli bir problem i\u00e7in en uygun veri yap\u0131s\u0131n\u0131 ve implementasyonunu se\u00e7me konusunda bilin\u00e7li kararlar vermeye \u00e7al\u0131\u015fmal\u0131s\u0131n\u0131z. Bu, geli\u015ftirici olarak yetkinli\u011finizin \u00f6nemli bir g\u00f6stergesidir.<\/p>\n Bu makalede, y\u0131\u011f\u0131n veri yap\u0131s\u0131n\u0131n derinliklerine inerek, temel prensibi olan LIFO'dan (Last-In, First-Out) ba\u015flay\u0131p, onu ger\u00e7ek d\u00fcnya senaryolar\u0131nda nas\u0131l kullan\u0131ld\u0131\u011f\u0131na, kendi y\u0131\u011f\u0131n\u0131m\u0131z\u0131 Python'da nas\u0131l kodlayabilece\u011fimize ve LeetCode gibi platformlardaki problem \u00e7\u00f6zme pratiklerine kadar geni\u015f bir perspektifte inceledik. Y\u0131\u011f\u0131n\u0131n sadece soyut bir kavram olmad\u0131\u011f\u0131n\u0131, aksine web taray\u0131c\u0131lar\u0131ndan metin d\u00fczenleyicilere, derleyicilerden i\u015fletim sistemlerinin bellek y\u00f6netimine kadar pek \u00e7ok alanda kritik bir rol oynayan temel bir yap\u0131 ta\u015f\u0131 oldu\u011funu g\u00f6rd\u00fck. \u00d6zellikle fonksiyon \u00e7a\u011fr\u0131 y\u0131\u011f\u0131n\u0131 \u00f6rne\u011fi, programlar\u0131n dahili i\u015fleyi\u015finde ne kadar merkezi bir rol \u00fcstlendi\u011fini g\u00f6zler \u00f6n\u00fcne serdi.<\/p>\n Kendi y\u0131\u011f\u0131n s\u0131n\u0131f\u0131m\u0131z\u0131 olu\u015ftururken, Unutmay\u0131n ki y\u0131\u011f\u0131n, veri yap\u0131lar\u0131 ve algoritmalar yolculu\u011funuzda kar\u015f\u0131la\u015faca\u011f\u0131n\u0131z ilk duraklardan sadece biridir. Ancak, bu temel yap\u0131y\u0131 sa\u011flam bir \u015fekilde kavramak, kuyruklar (queues), a\u011fa\u00e7lar (trees) ve graflar (graphs) gibi daha karma\u015f\u0131k veri yap\u0131lar\u0131n\u0131 anlamak i\u00e7in sa\u011flam bir zemin olu\u015fturacakt\u0131r. Bu bilgi birikimi, yaz\u0131l\u0131m m\u00fchendisli\u011fi m\u00fclakatlar\u0131nda ba\u015far\u0131l\u0131 olman\u0131zdan, daha verimli ve optimize edilmi\u015f kodlar yazman\u0131za kadar pek \u00e7ok alanda size avantaj sa\u011flayacakt\u0131r.<\/p>\n Gelecek ad\u0131mlar\u0131n\u0131z i\u00e7in \u00f6neriler:<\/p>\n Bu yolculukta edindi\u011finiz bilgilerle, yaz\u0131l\u0131m geli\u015ftirme d\u00fcnyas\u0131nda kar\u015f\u0131la\u015ft\u0131\u011f\u0131n\u0131z bir\u00e7ok problemi daha ak\u0131lc\u0131 ve verimli bir \u015fekilde \u00e7\u00f6zebilece\u011finizden eminiz. Ba\u015far\u0131lar!<\/p>\n Y\u0131\u011f\u0131n veri yap\u0131s\u0131 hakk\u0131nda akl\u0131n\u0131za tak\u0131labilecek baz\u0131 yayg\u0131n sorular\u0131 ve yan\u0131tlar\u0131n\u0131 a\u015fa\u011f\u0131da bulabilirsiniz:<\/p>\n C: Y\u0131\u011f\u0131n (Stack) \"Son Giren \u0130lk \u00c7\u0131kar\" (LIFO - Last-In, First-Out) prensibiyle \u00e7al\u0131\u015f\u0131r; yani en son eklenen eleman ilk \u00e7\u0131kar\u0131l\u0131r. Kuyruk (Queue) ise \"\u0130lk Giren \u0130lk \u00c7\u0131kar\" (FIFO - First-In, First-Out) prensibiyle \u00e7al\u0131\u015f\u0131r; yani ilk eklenen eleman ilk \u00e7\u0131kar\u0131l\u0131r.<\/p>\n<\/li>\n C: \"Stack Overflow\" hatas\u0131, program\u0131n \u00e7a\u011fr\u0131 y\u0131\u011f\u0131n\u0131n\u0131n (call stack) kapasitesini a\u015fmas\u0131 durumunda meydana gelir. Genellikle sonsuz \u00f6zyinelemeli (recursive) fonksiyon \u00e7a\u011fr\u0131lar\u0131 veya \u00e7ok derin fonksiyon \u00e7a\u011fr\u0131 zincirleri sonucunda bellek yetersizli\u011fi nedeniyle ortaya \u00e7\u0131kar.<\/p>\n<\/li>\n C: Y\u0131\u011f\u0131n\u0131n C: Teorik olarak ba\u015fka veri yap\u0131lar\u0131 (\u00f6rne\u011fin, bir dizi veya ba\u011fl\u0131 liste) kullan\u0131larak \"geri al\" i\u015flevi uygulanabilir ancak y\u0131\u011f\u0131n (LIFO prensibiyle) bu t\u00fcr bir i\u015flevsellik i\u00e7in en do\u011fal, sezgisel ve verimli veri yap\u0131s\u0131d\u0131r. Di\u011fer yakla\u015f\u0131mlar genellikle daha karma\u015f\u0131k olur veya ayn\u0131 performans\u0131 sa\u011flamayabilir.<\/p>\n<\/li>\n C: Mobil uygulamalarda navigasyon ak\u0131\u015f\u0131n\u0131 y\u00f6netmek i\u00e7in s\u0131k\u00e7a y\u0131\u011f\u0131n benzeri yap\u0131lar kullan\u0131l\u0131r. \u00d6rne\u011fin, bir kullan\u0131c\u0131n\u0131n bir dizi ekran aras\u0131nda gezinmesi (ekran push, geri tu\u015fu pop), uygulama i\u00e7i i\u015flemlerin ge\u00e7mi\u015fini y\u00f6netmek (undo\/redo) ve hatta baz\u0131 UI bile\u015fenlerinin durumu gibi senaryolarda y\u0131\u011f\u0131n mant\u0131\u011f\u0131ndan faydalan\u0131l\u0131r.<\/p>\n<\/li>\n<\/ul>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"Veri yap\u0131lar\u0131 ve algoritmalar d\u00fcnyas\u0131nda y\u0131\u011f\u0131n (stack), belki de en temel ama ayn\u0131 zamanda en g\u00fc\u00e7l\u00fc ara\u00e7lardan biridir.…","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"csco_page_header_type":"","csco_page_load_nextpost":"","csco_page_subscribe_form":"","csco_page_contact_form":"","footnotes":""},"categories":[1],"tags":[],"class_list":{"0":"post-33531","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-genel","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"\npush<\/code> ve pop<\/code>. Bu iki i\u015flem, y\u0131\u011f\u0131n\u0131n dinamik yap\u0131s\u0131n\u0131 olu\u015fturur ve eleman ekleme ile \u00e7\u0131karma mekanizmas\u0131n\u0131 y\u00f6netir.<\/p>\n\n
push<\/code> i\u015fleminde, yeni eleman y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcne konulur. \u00d6rne\u011fin, \u00fcst \u00fcste y\u0131\u011f\u0131lm\u0131\u015f tabaklar\u0131 d\u00fc\u015f\u00fcn\u00fcn; yeni bir tabak her zaman en \u00fcste yerle\u015ftirilir. Bu i\u015flem genellikle O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir, yani y\u0131\u011f\u0131n\u0131n boyutundan ba\u011f\u0131ms\u0131z olarak sabit bir s\u00fcrede ger\u00e7ekle\u015fir. \u00c7\u00fcnk\u00fc sadece y\u0131\u011f\u0131n\u0131n “\u00fcst” k\u0131sm\u0131nda bir de\u011fi\u015fiklik yap\u0131l\u0131r.<\/li>\nPop<\/code> i\u015flemi ise y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki eleman\u0131 \u00e7\u0131karmak ve d\u00f6nd\u00fcrmek i\u00e7in kullan\u0131l\u0131r. Taba\u011fa geri d\u00f6nersek, kirli bir tabak almak istedi\u011finizde her zaman en \u00fcsttekini al\u0131rs\u0131n\u0131z. E\u011fer y\u0131\u011f\u0131n bo\u015fsa ve pop<\/code> i\u015flemi \u00e7a\u011fr\u0131l\u0131rsa, bu bir hata (underflow) durumuna yol a\u00e7ar. Pop<\/code> i\u015flemi de genellikle O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir, \u00e7\u00fcnk\u00fc sadece en \u00fcstteki elemana eri\u015filir ve o eleman kald\u0131r\u0131l\u0131r.<\/li>\n<\/ul>\n\n
pop<\/code> veya peek<\/code> gibi i\u015flemleri \u00e7a\u011f\u0131rmadan \u00f6nce y\u0131\u011f\u0131n\u0131n bo\u015f olup olmad\u0131\u011f\u0131n\u0131 kontrol etmek, hatalar\u0131 \u00f6nlemek i\u00e7in olduk\u00e7a \u00f6nemlidir.<\/li>\nisFull<\/code> i\u015flemi y\u0131\u011f\u0131n\u0131n kapasitesine ula\u015f\u0131p ula\u015fmad\u0131\u011f\u0131n\u0131 kontrol eder. Bu durum genellikle dinamik olarak b\u00fcy\u00fcyebilen y\u0131\u011f\u0131n uygulamalar\u0131nda (\u00f6rne\u011fin Python listeleriyle) ge\u00e7erli de\u011fildir, ancak belle\u011fi k\u0131s\u0131tl\u0131 sistemlerde veya belirli s\u0131n\u0131rlamalarla \u00e7al\u0131\u015f\u0131ld\u0131\u011f\u0131nda \u00f6nem kazan\u0131r.<\/li>\nY\u0131\u011f\u0131n Nas\u0131l Uygulan\u0131r? (Diziler ve Ba\u011fl\u0131 Listeler)<\/h3>\n
Push<\/code> i\u015flemi: Dizinin sonuna yeni bir eleman ekleriz (\u00f6rne\u011fin, Python’da append()<\/code>, JavaScript’te push()<\/code>).
\n
\n Pop<\/code> i\u015flemi: Dizinin sonundan bir eleman\u0131 \u00e7\u0131kar\u0131r\u0131z (\u00f6rne\u011fin, Python’da pop()<\/code>, JavaScript’te pop()<\/code>).
\n
\n Peek<\/code> i\u015flemi: Dizinin son eleman\u0131na eri\u015firiz (\u00f6rne\u011fin, my_array[-1]<\/code>).<\/p>\nPush<\/code> i\u015flemi: Yeni bir d\u00fc\u011f\u00fcm (node) olu\u015fturulur ve listenin ba\u015f\u0131na eklenir. Bu, mevcut head<\/code> d\u00fc\u011f\u00fcm\u00fcn\u00fc yeni d\u00fc\u011f\u00fcm\u00fcn bir sonraki eleman\u0131 yapar ve yeni d\u00fc\u011f\u00fcm head<\/code> olur.
\n
\n Pop<\/code> i\u015flemi: Listenin ba\u015f\u0131ndaki d\u00fc\u011f\u00fcm kald\u0131r\u0131l\u0131r ve ba\u015f bir sonraki d\u00fc\u011f\u00fcme kayd\u0131r\u0131l\u0131r.
\n
\n Peek<\/code> i\u015flemi: Sadece head<\/code> d\u00fc\u011f\u00fcm\u00fcn\u00fcn de\u011ferine bak\u0131l\u0131r.<\/p>\nGer\u00e7ek D\u00fcnya Senaryolar\u0131: Y\u0131\u011f\u0131n Nerede Kar\u015f\u0131m\u0131za \u00c7\u0131kar? Pratik Kullan\u0131m Alanlar\u0131<\/h2>\n
Taray\u0131c\u0131 Ge\u00e7mi\u015fi ve Geri Alma Fonksiyonlar\u0131: Bir Vaka Analizi<\/h3>\n
history stack<\/code>) itilir. “Geri” (back) d\u00fc\u011fmesine bast\u0131\u011f\u0131n\u0131zda, taray\u0131c\u0131 y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki URL’yi \u00e7\u0131kar\u0131r ve sizi o sayfaya y\u00f6nlendirir. “\u0130leri” (forward) d\u00fc\u011fmesi ise genellikle ayr\u0131 bir y\u0131\u011f\u0131n veya ge\u00e7ici bir depolama mekanizmas\u0131 kullanarak, geri al\u0131nan sayfalar\u0131 tekrar y\u00fcklemeye olanak tan\u0131r. Bu sayede, internet \u00fczerinde sorunsuz bir \u015fekilde gezinebilir, \u00f6nceki sayfalara kolayca d\u00f6nebiliriz. Bu sistemde, her yeni sayfa y\u00fcklemesi bir push<\/code> i\u015flemi iken, geri tu\u015fu bir pop<\/code> i\u015flemidir.<\/p>\npush<\/code>) eklenir. Fonksiyon i\u015fini tamamlad\u0131\u011f\u0131nda, bu bilgiler y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131l\u0131r (pop<\/code>) ve kontrol, fonksiyondan \u00f6nceki konuma geri d\u00f6ner. Bu sayede, programlar i\u00e7 i\u00e7e fonksiyon \u00e7a\u011fr\u0131lar\u0131n\u0131, \u00f6zyinelemeli (recursive) algoritmalar\u0131 ve genel program ak\u0131\u015f\u0131n\u0131 hatas\u0131z bir \u015fekilde y\u00f6netebilir. E\u011fer bir program sonsuz bir \u00f6zyinelemeye girerse, \u00e7a\u011fr\u0131 y\u0131\u011f\u0131n\u0131 kapasitesini a\u015far ve me\u015fhur “stack overflow” hatas\u0131n\u0131 al\u0131rs\u0131n\u0131z. Bu hata, y\u0131\u011f\u0131n\u0131n sabit boyutlu oldu\u011fu durumlarda kapasite s\u0131n\u0131rlamalar\u0131na i\u015faret eder.<\/p>\n()<\/code>, {}<\/code>, []<\/code>) do\u011fru bir \u015fekilde e\u015fle\u015fip e\u015fle\u015fmedi\u011fini kontrol etmek i\u00e7in y\u0131\u011f\u0131nlar\u0131 kullan\u0131r. A\u00e7\u0131lan her parantez y\u0131\u011f\u0131na itilir. Kapanan bir parantez geldi\u011finde, y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki parantez \u00e7\u0131kar\u0131l\u0131r ve e\u015fle\u015fip e\u015fle\u015fmedi\u011fi kontrol edilir. E\u011fer e\u015fle\u015fme varsa devam edilir, yoksa veya y\u0131\u011f\u0131n bo\u015fsa bir hata bildirilir. Benzer \u015fekilde, aritmetik ifadelerin (infix, prefix, postfix) de\u011ferlendirilmesinde de y\u0131\u011f\u0131nlar kritik bir rol oynar. \u00d6rne\u011fin, postfix ifadeleri y\u0131\u011f\u0131n kullanarak kolayca de\u011ferlendirilebilir.<\/p>\nKendi Y\u0131\u011f\u0131n Yap\u0131m\u0131z\u0131 Kodlayal\u0131m: Ad\u0131m Ad\u0131m Python \u00d6rnekleri<\/h2>\n
Stack<\/code> ad\u0131nda bir s\u0131n\u0131f tan\u0131mlayarak ba\u015flayabiliriz. Bu s\u0131n\u0131f\u0131n yap\u0131c\u0131 metodunda (constructor) y\u0131\u011f\u0131n elemanlar\u0131n\u0131 tutacak bir Python listesi (array) ba\u015flataca\u011f\u0131z. Ard\u0131ndan, y\u0131\u011f\u0131n\u0131n temel i\u015flevlerini bu listeyi kullanarak uygulayaca\u011f\u0131z.<\/p>\nclass Stack:\n def __init__(self):\n # Y\u0131\u011f\u0131n elemanlar\u0131n\u0131 tutacak dahili bir Python listesi kullan\u0131yoruz.\n # Python listeleri dinamik diziler oldu\u011fu i\u00e7in boyut y\u00f6netimi bizim i\u00e7in otomatik yap\u0131l\u0131r.\n self.items = []\n\n def is_empty(self):\n # Y\u0131\u011f\u0131n\u0131n bo\u015f olup olmad\u0131\u011f\u0131n\u0131 kontrol eder.\n # Liste bo\u015fsa True, doluysa False d\u00f6nd\u00fcr\u00fcr.\n # return len(self.items) == 0 da kullan\u0131labilir.\n return not self.items\n\n def push(self, item):\n # Yeni bir eleman\u0131 y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcne (listenin sonuna) ekler.\n self.items.append(item)\n print(f\"'{item}' y\u0131\u011f\u0131na eklendi. Y\u0131\u011f\u0131n: {self.items}\")\n\n def pop(self):\n # Y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki eleman\u0131 \u00e7\u0131kar\u0131r ve d\u00f6nd\u00fcr\u00fcr.\n # E\u011fer y\u0131\u011f\u0131n bo\u015fsa, IndexError f\u0131rlat\u0131r\u0131z (underflow durumu).\n if not self.is_empty():\n removed_item = self.items.pop()\n print(f\"'{removed_item}' y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131ld\u0131. Y\u0131\u011f\u0131n: {self.items}\")\n return removed_item\n else:\n raise IndexError(\"Y\u0131\u011f\u0131n bo\u015f! \u00c7\u0131kar\u0131lacak eleman yok.\")\n\n def peek(self, n=1):\n # Y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki eleman\u0131 (veya n'inci eleman\u0131) g\u00f6sterir, \u00e7\u0131karmadan.\n # n=1 varsay\u0131lan olarak en \u00fcstteki eleman\u0131 g\u00f6sterir.\n if not self.is_empty():\n if n > 0 and n <= len(self.items):\n # -n indeksi sondan n'inci eleman\u0131 verir.\n print(f\"Y\u0131\u011f\u0131n\u0131n \u00fcst\u00fcnden {n}. eleman: '{self.items[-n]}'\")\n return self.items[-n]\n else:\n raise IndexError(f\"Ge\u00e7ersiz n de\u011feri veya yeterli eleman yok. Y\u0131\u011f\u0131n boyutu: {self.size()}\")\n else:\n raise IndexError(\"Y\u0131\u011f\u0131n bo\u015f! G\u00f6sterilecek eleman yok.\")\n\n def size(self):\n # Y\u0131\u011f\u0131ndaki eleman say\u0131s\u0131n\u0131 d\u00f6nd\u00fcr\u00fcr.\n current_size = len(self.items)\n print(f\"Y\u0131\u011f\u0131n\u0131n mevcut boyutu: {current_size}\")\n return current_size\n\n def display(self):\n # Y\u0131\u011f\u0131n\u0131n mevcut durumunu g\u00f6sterir.\n if self.is_empty():\n print(\"Y\u0131\u011f\u0131n bo\u015f.\")\n else:\n print(f\"Y\u0131\u011f\u0131n i\u00e7eri\u011fi (\u00fcstten alta): {self.items[::-1]}\") # Ters \u00e7evirerek LIFO s\u0131ras\u0131n\u0131 vurguluyoruz\n<\/pre>\nStack<\/code> s\u0131n\u0131f\u0131m\u0131z\u0131n t\u00fcm temel i\u015flevlerini i\u00e7ermektedir. __init__<\/code> metodunda self.items = []<\/code> ile bo\u015f bir liste ba\u015flat\u0131yoruz. is_empty<\/code> metodu, listenin bo\u015f olup olmad\u0131\u011f\u0131n\u0131 kontrol eder. push<\/code> metodu, append()<\/code> fonksiyonu ile listenin sonuna (y\u0131\u011f\u0131n\u0131n \u00fcst\u00fcne) eleman ekler. pop<\/code> metodu ise pop()<\/code> fonksiyonu ile listenin sonundaki eleman\u0131 \u00e7\u0131kar\u0131r ve d\u00f6nd\u00fcr\u00fcr. peek<\/code> metodu, listenin son eleman\u0131na [-1]<\/code> indeksi ile eri\u015ferek y\u0131\u011f\u0131n\u0131n en \u00fcst eleman\u0131n\u0131 g\u00f6sterir. size<\/code> metodu ise len()<\/code> fonksiyonu ile liste boyutunu d\u00f6nd\u00fcr\u00fcr. display<\/code> metodu ise y\u0131\u011f\u0131n\u0131n o anki durumunu g\u00f6rselle\u015ftirmeye yard\u0131mc\u0131 olur.<\/p>\nStack<\/code> s\u0131n\u0131f\u0131n\u0131 kullanarak nas\u0131l i\u015flemler yapabilece\u011fimize dair bir \u00f6rnek senaryoya bakal\u0131m:<\/p>\n# Bir y\u0131\u011f\u0131n nesnesi olu\u015ftural\u0131m\nmy_stack = Stack()\n\n# Y\u0131\u011f\u0131n\u0131n ba\u015flang\u0131\u00e7 durumunu kontrol edelim\nmy_stack.is_empty() # \u00c7\u0131kt\u0131: Y\u0131\u011f\u0131n\u0131n mevcut boyutu: 0, True\n\n# Elemanlar\u0131 y\u0131\u011f\u0131na ekleyelim (Push i\u015flemleri)\nmy_stack.push(\"Kitap A\")\nmy_stack.push(\"Kitap B\")\nmy_stack.push(\"Kitap C\")\n\n# Y\u0131\u011f\u0131n\u0131n mevcut durumunu ve boyutunu kontrol edelim\nmy_stack.display() # \u00c7\u0131kt\u0131: Y\u0131\u011f\u0131n i\u00e7eri\u011fi (\u00fcstten alta): ['Kitap C', 'Kitap B', 'Kitap A']\nmy_stack.size() # \u00c7\u0131kt\u0131: Y\u0131\u011f\u0131n\u0131n mevcut boyutu: 3, 3\n\n# Y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki elemana bakal\u0131m (Peek i\u015flemi)\nmy_stack.peek() # \u00c7\u0131kt\u0131: Y\u0131\u011f\u0131n\u0131n \u00fcst\u00fcndeki eleman: 'Kitap C', 'Kitap C'\nmy_stack.peek(2) # \u00c7\u0131kt\u0131: Y\u0131\u011f\u0131n\u0131n \u00fcst\u00fcnden 2. eleman: 'Kitap B', 'Kitap B'\n\n# Elemanlar\u0131 y\u0131\u011f\u0131ndan \u00e7\u0131karal\u0131m (Pop i\u015flemleri)\nmy_stack.pop() # \u00c7\u0131kt\u0131: 'Kitap C' y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131ld\u0131. Y\u0131\u011f\u0131n: ['Kitap A', 'Kitap B']\nmy_stack.pop() # \u00c7\u0131kt\u0131: 'Kitap B' y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131ld\u0131. Y\u0131\u011f\u0131n: ['Kitap A']\n\n# Y\u0131\u011f\u0131n\u0131n bo\u015f olup olmad\u0131\u011f\u0131n\u0131 tekrar kontrol edelim\nmy_stack.is_empty() # \u00c7\u0131kt\u0131: False (\u00e7\u00fcnk\u00fc hala 'Kitap A' var)\n\n# Kalan son eleman\u0131 da \u00e7\u0131karal\u0131m\nmy_stack.pop() # \u00c7\u0131kt\u0131: 'Kitap A' y\u0131\u011f\u0131ndan \u00e7\u0131kar\u0131ld\u0131. Y\u0131\u011f\u0131n: []\n\n# \u015eimdi y\u0131\u011f\u0131n bo\u015f olmal\u0131\nmy_stack.is_empty() # \u00c7\u0131kt\u0131: True\n\n# Bo\u015f bir y\u0131\u011f\u0131ndan pop yapmaya \u00e7al\u0131\u015f\u0131rsak ne olur?\ntry:\n my_stack.pop()\nexcept IndexError as e:\n print(f\"Hata yakaland\u0131: {e}\") # \u00c7\u0131kt\u0131: Hata yakaland\u0131: Y\u0131\u011f\u0131n bo\u015f! \u00c7\u0131kar\u0131lacak eleman yok.\n\n# Bo\u015f bir y\u0131\u011f\u0131ndan peek yapmaya \u00e7al\u0131\u015f\u0131rsak ne olur?\ntry:\n my_stack.peek()\nexcept IndexError as e:\n print(f\"Hata yakaland\u0131: {e}\") # \u00c7\u0131kt\u0131: Hata yakaland\u0131: Y\u0131\u011f\u0131n bo\u015f! G\u00f6sterilecek eleman yok.\n<\/pre>\nmy_stack<\/code> ad\u0131nda bir y\u0131\u011f\u0131n olu\u015fturduk, \u00e7e\u015fitli elemanlar ekledik, y\u0131\u011f\u0131n\u0131n durumunu kontrol ettik ve elemanlar\u0131 \u00e7\u0131kard\u0131k. Ayr\u0131ca, bo\u015f bir y\u0131\u011f\u0131n \u00fczerinde pop<\/code> ve peek<\/code> yapmaya \u00e7al\u0131\u015farak nas\u0131l hata y\u00f6netimi yap\u0131ld\u0131\u011f\u0131n\u0131 da g\u00f6sterdik. Python'\u0131n yerle\u015fik listeleri sayesinde, bu t\u00fcr bir veri yap\u0131s\u0131n\u0131 uygulamak olduk\u00e7a sezgisel ve basittir. Ancak bu basitlik, y\u0131\u011f\u0131n\u0131n alt\u0131nda yatan temel prensipleri ve LIFO mant\u0131\u011f\u0131n\u0131 anlamak i\u00e7in bir engel te\u015fkil etmemelidir. \u00d6nemli olan, bu yap\u0131lar\u0131n nas\u0131l \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131 ve hangi durumlarda kullan\u0131labilece\u011fini kavramakt\u0131r.<\/p>\nLeetCode Prati\u011fi: Y\u0131\u011f\u0131n Temelli Algoritmalar ve Problemler<\/h2>\n
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()<\/code>, {}<\/code>, []<\/code>) do\u011fru s\u0131rada ve e\u015fle\u015fmi\u015f olup olmad\u0131\u011f\u0131n\u0131 kontrol etme.<\/li>\nGe\u00e7erli Parantezler Problemi (Valid Parentheses): Y\u0131\u011f\u0131nla \u00c7\u00f6z\u00fcm<\/h3>\n
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push<\/code> ederiz.<\/li>\n\n
pop<\/code> ederiz. Bu eleman, mevcut kapan\u0131\u015f parantezimizle e\u015fle\u015fen a\u00e7\u0131l\u0131\u015f parantezi olmal\u0131d\u0131r. E\u015fle\u015fme yoksa (\u00f6rne\u011fin, ')' i\u00e7in '{' \u00e7\u0131kt\u0131ysa), string ge\u00e7ersizdir.<\/li>\n<\/ul>\n<\/li>\ndef is_valid_parentheses(s: str) -> bool:\n stack = [] # Y\u0131\u011f\u0131n\u0131m\u0131z\u0131 Python listesi olarak kullan\u0131yoruz\n # Parantez e\u015fle\u015fmelerini kolayca kontrol etmek i\u00e7in bir s\u00f6zl\u00fck tan\u0131ml\u0131yoruz\n mapping = {\")\": \"(\", \"}\": \"{\", \"]\": \"[\"}\n\n for char in s:\n if char in mapping: # E\u011fer karakter bir kapan\u0131\u015f parantezi ise\n # Y\u0131\u011f\u0131n bo\u015fsa veya y\u0131\u011f\u0131n\u0131n en \u00fcst\u00fcndeki eleman kapan\u0131\u015f parantezinin e\u015fi de\u011filse\n top_element = stack.pop() if stack else '#' # Y\u0131\u011f\u0131n bo\u015fsa '#' koy ki e\u015fle\u015fme olmas\u0131n\n if mapping[char] != top_element:\n return False\n else: # E\u011fer karakter bir a\u00e7\u0131l\u0131\u015f parantezi ise\n stack.append(char) # Y\u0131\u011f\u0131na ekle\n\n # T\u00fcm stringi i\u015fledikten sonra y\u0131\u011f\u0131n bo\u015fsa ge\u00e7erlidir, aksi halde de\u011fil\n return not stack\n\n# Test \u00f6rnekleri\nprint(f\"'()' ge\u00e7erli mi? {is_valid_parentheses('()')}\") # True\nprint(f\"'()[]{{}}' ge\u00e7erli mi? {is_valid_parentheses('()[]{}')}\") # True\nprint(f\"'(]' ge\u00e7erli mi? {is_valid_parentheses('(]')}\") # False\nprint(f\"'([)]' ge\u00e7erli mi? {is_valid_parentheses('([)]')}\") # False\nprint(f\"'{{[]}}' ge\u00e7erli mi? {is_valid_parentheses('{[()]}}')}\") # True\nprint(f\"'' ge\u00e7erli mi? {is_valid_parentheses('')}\") # True\nprint(f\"'[' ge\u00e7erli mi? {is_valid_parentheses('[')}\") # False\n<\/pre>\n\u0130leri D\u00fczey \u0130pu\u00e7lar\u0131: Y\u0131\u011f\u0131n Performans\u0131n\u0131 Optimize Etmek ve S\u0131n\u0131r Durumlar<\/h2>\n
Zaman ve Yer Karma\u015f\u0131kl\u0131\u011f\u0131 Analizi<\/h3>\n
push<\/code>, pop<\/code> ve peek<\/code>, ideal durumda genellikle O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir. Yani, y\u0131\u011f\u0131n\u0131n boyutu ne kadar b\u00fcy\u00fck olursa olsun, bu i\u015flemler sabit bir s\u00fcrede tamamlan\u0131r. Bu, y\u0131\u011f\u0131n\u0131n bir\u00e7ok uygulama i\u00e7in tercih edilmesinin ana nedenlerinden biridir. Yer karma\u015f\u0131kl\u0131\u011f\u0131 ise, y\u0131\u011f\u0131nda depolanan eleman say\u0131s\u0131 kadar olacakt\u0131r, yani O(N), N eleman say\u0131s\u0131d\u0131r.<\/p>\nS\u0131n\u0131r Durumlar\u0131 ve Hata Y\u00f6netimi<\/h3>\n
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Pop<\/code> veya Peek<\/code>:<\/strong> Y\u0131\u011f\u0131n bo\u015fken bir eleman\u0131 \u00e7\u0131karmaya (pop<\/code>) veya en \u00fcstteki elemana bakmaya (peek<\/code>) \u00e7al\u0131\u015fmak \"underflow\" durumuna yol a\u00e7ar. Bu durumda bir hata (\u00f6rne\u011fin, Python'da IndexError<\/code>) f\u0131rlatmak veya \u00f6zel bir de\u011fer (None<\/code>, null<\/code> vb.) d\u00f6nd\u00fcrmek uygun bir hata y\u00f6netimi stratejisidir. Kod \u00f6rneklerimizde bu durumu IndexError<\/code> ile y\u00f6nettik.<\/li>\nPush<\/code> (Sabit Boyutlu Y\u0131\u011f\u0131nlar \u0130\u00e7in):<\/strong> E\u011fer y\u0131\u011f\u0131n sabit bir boyuta sahip bir diziyle uygulanm\u0131\u015fsa ve kapasitesi doluyken yeni bir eleman eklemeye (push<\/code>) \u00e7al\u0131\u015f\u0131l\u0131rsa, bu \"overflow\" durumuna yol a\u00e7ar. Bu durumda da bir hata f\u0131rlat\u0131lmas\u0131 veya eleman\u0131n eklenemedi\u011fini belirten bir d\u00f6n\u00fc\u015f de\u011feri sa\u011flanmas\u0131 gerekir. Dinamik diziler (Python listeleri) bu sorunu otomatik olarak y\u00f6netti\u011fi i\u00e7in bu durum Python'da do\u011frudan bir problem yaratmaz, ancak d\u00fc\u015f\u00fck seviyeli dillerde veya belle\u011fi k\u0131s\u0131tl\u0131 ortamlarda kritik bir konudur.<\/li>\npop<\/code> veya peek<\/code> i\u015flemleri. Bu durum, y\u0131\u011f\u0131n\u0131n bo\u015f olma durumuna ge\u00e7i\u015fini do\u011fru bir \u015fekilde y\u00f6netmeyi gerektirir.<\/li>\n<\/ul>\nOptimize Edilmi\u015f Y\u0131\u011f\u0131n Tasar\u0131mlar\u0131<\/h3>\n
Sonu\u00e7: Y\u0131\u011f\u0131n\u0131n G\u00fcc\u00fc ve Gelecekteki Ad\u0131mlar\u0131n\u0131z<\/h2>\n
push<\/code>, pop<\/code>, peek<\/code>, isEmpty<\/code> ve size<\/code> gibi temel i\u015flemlerin nas\u0131l \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131 ad\u0131m ad\u0131m Python kod \u00f6rnekleriyle peki\u015ftirdik. Bu pratik uygulama, teorik bilgileri somutla\u015ft\u0131rmak ve algoritmik d\u00fc\u015f\u00fcnme becerilerinizi geli\u015ftirmek i\u00e7in paha bi\u00e7ilmez bir ad\u0131md\u0131. Ayr\u0131ca, \"Ge\u00e7erli Parantezler\" problemi \u00fczerinden y\u0131\u011f\u0131n\u0131n problem \u00e7\u00f6zme yetene\u011fini ke\u015ffederken, LeetCode'daki benzer sorunlara nas\u0131l yakla\u015fman\u0131z gerekti\u011fi konusunda bir k\u0131lavuz edindiniz.<\/p>\n\n
S\u0131k\u00e7a Sorulan Sorular: Y\u0131\u011f\u0131n Hakk\u0131nda Merak Edilenler<\/h2>\n
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push<\/code>, pop<\/code> ve peek<\/code> gibi temel i\u015flemleri, elemanlar\u0131n her zaman y\u0131\u011f\u0131n\u0131n bir ucundan (\u00fcst\u00fcnden) eklenip \u00e7\u0131kar\u0131lmas\u0131yla ilgilidir. Bu, y\u0131\u011f\u0131n\u0131n boyutundan ba\u011f\u0131ms\u0131z olarak sabit bir say\u0131da i\u015flem gerektirdi\u011fi i\u00e7in O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir. Ancak dinamik dizi tabanl\u0131 uygulamalarda ara s\u0131ra ger\u00e7ekle\u015fen yeniden boyutland\u0131rma i\u015flemleri ge\u00e7ici olarak O(N) maliyetine neden olabilir, ancak amortize edilmi\u015f maliyet yine de O(1)'dir.<\/p>\n<\/li>\n