{"id":7644,"date":"2024-12-25T14:30:10","date_gmt":"2024-12-25T11:30:10","guid":{"rendered":"https:\/\/fatihsoysal.com\/blog\/leetcode-zikzak-donusumu-javascript-cozumu\/"},"modified":"2024-12-25T14:30:10","modified_gmt":"2024-12-25T11:30:10","slug":"leetcode-zikzak-donusumu-javascript-cozumu","status":"publish","type":"post","link":"https:\/\/fatihsoysal.com\/blog\/leetcode-zikzak-donusumu-javascript-cozumu\/","title":{"rendered":"LeetCode Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc: JavaScript \u00c7\u00f6z\u00fcm\u00fc"},"content":{"rendered":"<p>LeetCode Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc: JavaScript \u00c7\u00f6z\u00fcm\u00fc<\/p>\n<p>Merhaba, Fatih Soysal burada! Bu yaz\u0131m\u0131zda, LeetCode&#8217;un pop\u00fcler problemlerinden biri olan &#8220;Zigzag Conversion&#8221; (Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc) problemine JavaScript ile nas\u0131l \u00e7\u00f6z\u00fcm getirilebilece\u011fini detayl\u0131 bir \u015fekilde ele alaca\u011f\u0131z.  Bu problem, verilen bir metni zikzak \u015feklinde d\u00fczenleyerek yeni bir metin olu\u015fturmam\u0131z\u0131 gerektirir.  \u00d6ncelikle problemi daha iyi anlamak i\u00e7in bir \u00f6rne\u011fe bakal\u0131m.<\/p>\n<h2>Problem Tan\u0131m\u0131<\/h2>\n<p>Verilen bir metin ve sat\u0131r say\u0131s\u0131 ile, metni zikzak \u015feklinde d\u00fczenleyerek yeni bir metin olu\u015fturmam\u0131z gerekiyor. \u00d6rne\u011fin, &#8220;PAYPALISHIRING&#8221; metnini 3 sat\u0131rda zikzak d\u00fczenledi\u011fimizde sonu\u00e7 &#8220;PAHNAPLSIIGYIR&#8221; olacakt\u0131r.  Bu nas\u0131l oluyor?  \u0130lk sat\u0131ra &#8216;P&#8217;, &#8216;A&#8217;, &#8216;H&#8217;, &#8216;N&#8217;, &#8216;A&#8217;, &#8216;P&#8217;, &#8216;I&#8217; yaz\u0131l\u0131r, ikinci sat\u0131ra &#8216;Y&#8217;, &#8216;L&#8217;, &#8216;S&#8217;, &#8216;I&#8217;, &#8216;G&#8217; yaz\u0131l\u0131r, \u00fc\u00e7\u00fcnc\u00fc sat\u0131ra ise &#8216;I&#8217;, &#8216;R&#8217; yaz\u0131l\u0131r. Sonra bu sat\u0131rlar birle\u015ftirilerek yeni metin elde edilir. <\/p>\n<p>Problemin zorlu\u011fu, bu zikzak d\u00fczenini programatik olarak olu\u015fturmakt\u0131r.  Bu noktada, algoritmik d\u00fc\u015f\u00fcnme becerilerimizi kullanmam\u0131z gerekiyor.  \u015eimdi gelin, bu problemi nas\u0131l \u00e7\u00f6zebilece\u011fimize bakal\u0131m. <\/p>\n<h2>JavaScript ile Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc \u00c7\u00f6z\u00fcm\u00fc<\/h2>\n<p>Bu problemi \u00e7\u00f6zmek i\u00e7in birka\u00e7 farkl\u0131 yakla\u015f\u0131m kullan\u0131labilir.  Ancak, en verimli ve okunabilir y\u00f6ntemlerden biri,  dizilerin ve d\u00f6ng\u00fclerin etkili bir \u015fekilde kullan\u0131lmas\u0131na dayan\u0131r.  A\u015fa\u011f\u0131daki JavaScript kodunda, bu y\u00f6ntemi g\u00f6rebilirsiniz:<\/p>\n<pre><code class=\"language-javascript\">\n\/**\n * @param {string} s\n * @param {number} numRows\n * @return {string}\n *\/\nvar convert = function(s, numRows) {\n    if (numRows === 1) return s; \/\/ E\u011fer tek sat\u0131r ise d\u00f6n\u00fc\u015f\u00fcm gerekmiyor\n\n    const result = new Array(numRows).fill('');\n    let row = 0;\n    let down = true;\n\n    for (let i = 0; i < s.length; i++) {\n        result[row] += s[i];\n        if (down) {\n            row++;\n            if (row === numRows) {\n                row -= 2;\n                down = false;\n            }\n        } else {\n            row--;\n            if (row === -1) {\n                row += 2;\n                down = true;\n            }\n        }\n    }\n\n    return result.join('');\n};\n<\/code><\/pre>\n<p>Bu kod, \u00f6ncelikle numRows de\u011fi\u015fkenini kontrol eder. E\u011fer numRows 1 ise,  zikzak d\u00f6n\u00fc\u015f\u00fcm\u00fcne gerek yoktur ve orijinal metin d\u00f6nd\u00fcr\u00fcl\u00fcr. Aksi takdirde, result ad\u0131nda bir dizi olu\u015fturulur ve bu dizi, her bir sat\u0131r\u0131 temsil eder.  Ard\u0131ndan, bir d\u00f6ng\u00fc ile metin karakterleri i\u015flenir ve her karakter do\u011fru sat\u0131ra eklenir.  down de\u011fi\u015fkeni, karakterin a\u015fa\u011f\u0131 m\u0131 yoksa yukar\u0131 m\u0131 hareket etti\u011fini takip eder. Son olarak,  result dizisinin elemanlar\u0131 birle\u015ftirilerek zikzak d\u00f6n\u00fc\u015ft\u00fcr\u00fclm\u00fc\u015f metin elde edilir.<\/p>\n<h2>Performans Analizi ve Optimizasyonlar<\/h2>\n<p>Yukar\u0131daki kod, zaman karma\u015f\u0131kl\u0131\u011f\u0131 O(n) olan do\u011frusal bir zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir, burada n metnin uzunlu\u011fudur.  Bu, olduk\u00e7a verimli bir \u00e7\u00f6z\u00fcmd\u00fcr.  Bellek karma\u015f\u0131kl\u0131\u011f\u0131 da O(n) seviyesindedir.  Daha fazla optimizasyon i\u00e7in, \u00f6zellikle \u00e7ok b\u00fcy\u00fck metinler i\u00e7in,  uzun metinleri par\u00e7alara ay\u0131r\u0131p i\u015flemek gibi stratejiler d\u00fc\u015f\u00fcn\u00fclebilir. Ancak bu problem i\u00e7in mevcut \u00e7\u00f6z\u00fcm yeterli performans g\u00f6sterir.<\/p>\n<h2>Sonu\u00e7<\/h2>\n<p>Bu yaz\u0131da, LeetCode'un \"Zigzag Conversion\" problemine JavaScript ile verimli bir \u00e7\u00f6z\u00fcm sunmay\u0131 ama\u00e7lad\u0131k.  Umar\u0131m bu makale,  hem problemi anlaman\u0131za hem de  JavaScript ile etkili kod yazman\u0131za yard\u0131mc\u0131 olmu\u015ftur.  Daha fazla bilgi ve kaynak i\u00e7in  <a href=\"https:\/\/fatihsoysal.com\" target=\"_blank\" rel=\"noopener noreferrer\">fatihsoysal.com<\/a> adresini ziyaret edebilirsiniz.  Ayr\u0131ca, bu t\u00fcr problemlerle daha fazla ilgileniyorsan\u0131z,  <a href=\"https:\/\/dev.to\/rahulgithubweb\/leetcode-challenge-6-zigzag-conversion-javascript-solution-4lac\" target=\"_blank\" rel=\"noopener noreferrer\">buradaki<\/a>  \u0130ngilizce makaleden de faydalanabilirsiniz.  Kodlama yolculu\u011funuza devam ederken ba\u015far\u0131lar dilerim!<\/p>\n<p><strong>#Etiketler:<\/strong> LeetCode, Zigzag Conversion, JavaScript, Algoritma, Programlama, Kodlama, Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc,  Problem \u00c7\u00f6zme,  Web Geli\u015ftirme,  Fatih Soysal<\/p>\n","protected":false},"excerpt":{"rendered":"LeetCode Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc: JavaScript \u00c7\u00f6z\u00fcm\u00fc Merhaba, Fatih Soysal burada! Bu yaz\u0131m\u0131zda, LeetCode&#8217;un pop\u00fcler problemlerinden biri olan &#8220;Zigzag Conversion&#8221;&hellip;","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-7644","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-genel","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v20.5 (Yoast SEO v25.3.1) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>LeetCode Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc: JavaScript \u00c7\u00f6z\u00fcm\u00fc<\/title>\n<meta name=\"description\" content=\"Merhaba, Fatih Soysal burada! 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Bu problem, verilen bir metni zikzak \u015feklinde d\u00fczenleyerek yeni bir metin olu\u015fturmam\u0131z\u0131 gerektirir. \u00d6ncelikle problemi daha iyi anlamak i\u00e7in bir \u00f6rne\u011fe bakal\u0131m.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/fatihsoysal.com\/blog\/leetcode-zikzak-donusumu-javascript-cozumu\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"LeetCode Zikzak D\u00f6n\u00fc\u015f\u00fcm\u00fc: JavaScript \u00c7\u00f6z\u00fcm\u00fc\" \/>\n<meta property=\"og:description\" content=\"Merhaba, Fatih Soysal burada! 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