LeetCode 148: Ba\u011flant\u0131l\u0131 Listeyi S\u0131ralama<\/p>\n
Merhaba, ben Fatih Soysal. Bu makalede, LeetCode’un 148. problemi olan “Ba\u011flant\u0131l\u0131 Listeyi S\u0131ralama” problemini detayl\u0131 bir \u015fekilde ele alaca\u011f\u0131z. Problem, verilen bir ba\u011flant\u0131l\u0131 listeyi artan s\u0131rada s\u0131ralamam\u0131z\u0131 istiyor. Bu problemi \u00e7\u00f6zmek i\u00e7in birka\u00e7 farkl\u0131 yakla\u015f\u0131m mevcuttur; ancak en verimli y\u00f6ntemlerden biri Merge Sort algoritmas\u0131d\u0131r. \u00d6ncelikle problemi daha iyi anlamak i\u00e7in, problemin detaylar\u0131na ve Merge Sort algoritmas\u0131n\u0131n nas\u0131l \u00e7al\u0131\u015ft\u0131\u011f\u0131na bakal\u0131m.<\/p>\n
Verilen bir ba\u011flant\u0131l\u0131 listeyi, elemanlar\u0131n\u0131 artan s\u0131rada s\u0131ralayarak yeni bir ba\u011flant\u0131l\u0131 liste olu\u015fturmam\u0131z gerekiyor. Bu i\u015flem s\u0131ras\u0131nda orijinal ba\u011flant\u0131l\u0131 listeyi de\u011fi\u015ftirmemeliyiz. \u00d6rne\u011fin, 4->2->1->3 gibi bir listeyi 1->2->3->4 \u015feklinde s\u0131ralamam\u0131z gerekiyor.<\/p>\n
Merge Sort, b\u00f6l ve y\u00f6net (divide and conquer) yakla\u015f\u0131m\u0131 kullanan bir s\u0131ralama algoritmas\u0131d\u0131r. Bu algoritma, listeyi s\u00fcrekli olarak ikiye b\u00f6lerek, her bir alt listenin s\u0131ral\u0131 olmas\u0131n\u0131 sa\u011flar. Daha sonra, bu s\u0131ral\u0131 alt listeleri birle\u015ftirerek, t\u00fcm listenin s\u0131ral\u0131 halini elde ederiz. Bu y\u00f6ntem, ba\u011flant\u0131l\u0131 listeler i\u00e7in olduk\u00e7a uygundur, \u00e7\u00fcnk\u00fc ba\u011flant\u0131l\u0131 listelerde elemanlar\u0131n rastgele eri\u015fimi yava\u015ft\u0131r.<\/p>\n
Algoritman\u0131n ad\u0131mlar\u0131 \u015funlard\u0131r:<\/strong><\/p>\n Bu kod, Merge Sort algoritmas\u0131n\u0131 kullanarak ba\u011flant\u0131l\u0131 listeyi s\u0131ralar. getMid fonksiyonu listenin ortas\u0131n\u0131 bulur, merge fonksiyonu ise iki s\u0131ral\u0131 listeyi birle\u015ftirir. Sonu\u00e7 olarak, s\u0131ralama i\u015flemi logn zaman karma\u015f\u0131kl\u0131\u011f\u0131 ile ger\u00e7ekle\u015ftirilir.<\/p>\n Bu problem, di\u011fer s\u0131ralama algoritmalar\u0131 ile de \u00e7\u00f6z\u00fclebilir. Ancak, Merge Sort, ba\u011flant\u0131l\u0131 listeler i\u00e7in en uygun algoritmalardan biridir. Di\u011fer algoritmalar, ba\u011flant\u0131l\u0131 listelerin yap\u0131s\u0131 nedeniyle daha d\u00fc\u015f\u00fck performans g\u00f6sterebilir.<\/p>\n Ba\u015fka bir \u00e7\u00f6z\u00fcm olarak, ba\u011flant\u0131l\u0131 listeyi \u00f6nce bir diziye aktarabilir, diziyi s\u0131ralamak i\u00e7in h\u0131zl\u0131 bir algoritma kullanabilir ve ard\u0131ndan s\u0131ralama sonucundan yeni bir ba\u011flant\u0131l\u0131 liste olu\u015fturabilirsiniz. Ancak bu yakla\u015f\u0131m, ek bellek gerektirir.<\/p>\n LeetCode 148 problemi, ba\u011flant\u0131l\u0131 liste s\u0131ralama konusunda \u00f6nemli bir problemdir. Merge Sort algoritmas\u0131, bu problem i\u00e7in verimli ve etkili bir \u00e7\u00f6z\u00fcm sunmaktad\u0131r. Umar\u0131m bu makale, problemi anlaman\u0131za ve \u00e7\u00f6zmenize yard\u0131mc\u0131 olmu\u015ftur. Daha fazla bilgi i\u00e7in fatihsoysal.com<\/a> sitesini ziyaret edebilirsiniz. Ayr\u0131ca, konu hakk\u0131nda daha fazla bilgi edinmek i\u00e7in bu linki<\/a> inceleyebilirsiniz. <\/p>\n #LeetCode #148 #Ba\u011flant\u0131l\u0131Liste #S\u0131ralama #MergeSort #Algoritma #Programlama #Java #VeriYap\u0131lar\u0131 #FatihSoysal<\/p>\n <\/body>\n
Kod \u00d6rne\u011fi (Java)<\/h2>\n
\npublic ListNode sortList(ListNode head) {\n if (head == null || head.next == null) {\n return head;\n }\n ListNode mid = getMid(head);\n ListNode right = sortList(mid);\n mid.next = null;\n ListNode left = sortList(head);\n return merge(left, right);\n}\n\nprivate ListNode getMid(ListNode head) {\n ListNode slow = head, fast = head.next;\n while (fast != null && fast.next != null) {\n slow = slow.next;\n fast = fast.next.next;\n }\n return slow;\n}\n\nprivate ListNode merge(ListNode l1, ListNode l2) {\n ListNode dummy = new ListNode(0);\n ListNode tail = dummy;\n while (l1 != null && l2 != null) {\n if (l1.val < l2.val) {\n tail.next = l1;\n l1 = l1.next;\n } else {\n tail.next = l2;\n l2 = l2.next;\n }\n tail = tail.next;\n }\n tail.next = (l1 != null) ? l1 : l2;\n return dummy.next;\n}\n<\/code><\/pre>\nAlternatif \u00c7\u00f6z\u00fcmler<\/h2>\n
Sonu\u00e7<\/h2>\n
#Etiketler<\/h3>\n
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