{"id":8398,"date":"2025-01-02T11:00:17","date_gmt":"2025-01-02T08:00:17","guid":{"rendered":"https:\/\/fatihsoysal.com\/blog\/leetcode-15-3sum-javascript-cozumu\/"},"modified":"2025-01-02T11:00:17","modified_gmt":"2025-01-02T08:00:17","slug":"leetcode-15-3sum-javascript-cozumu","status":"publish","type":"post","link":"https:\/\/fatihsoysal.com\/blog\/leetcode-15-3sum-javascript-cozumu\/","title":{"rendered":"LeetCode 15: 3Sum JavaScript \u00c7\u00f6z\u00fcm\u00fc"},"content":{"rendered":"

LeetCode 15: 3Sum JavaScript \u00c7\u00f6z\u00fcm\u00fc<\/p>\n

Bu makalede, LeetCode’un pop\u00fcler problemlerinden biri olan “3Sum” problemine JavaScript ile nas\u0131l \u00e7\u00f6z\u00fcm \u00fcretebilece\u011finizi detayl\u0131 olarak ele alaca\u011f\u0131z. Hedefimiz, verilen bir dizide toplam\u0131 s\u0131f\u0131ra e\u015fit olan \u00fc\u00e7 say\u0131n\u0131n t\u00fcm olas\u0131 kombinasyonlar\u0131n\u0131 bulmak. Bu problem, algoritma ve veri yap\u0131lar\u0131 bilginizi test eden zorlu bir problemdir. Ancak do\u011fru yakla\u015f\u0131mla, verimli bir \u00e7\u00f6z\u00fcm geli\u015ftirmek m\u00fcmk\u00fcnd\u00fcr.<\/p>\n

Problemin Anlat\u0131m\u0131<\/h2>\n

Verilen bir tam say\u0131 dizisi nums<\/code>, toplam\u0131 s\u0131f\u0131ra e\u015fit olan \u00fc\u00e7 say\u0131n\u0131n t\u00fcm olas\u0131, benzersiz kombinasyonlar\u0131n\u0131 bulun. Her kombinasyon, artan s\u0131rada listelenmelidir. \u00d6rne\u011fin, nums = [-1, 0, 1, 2, -1, -4]<\/code> i\u00e7in, \u00e7\u0131kt\u0131 [[-1, -1, 2], [-1, 0, 1]]<\/code> olmal\u0131d\u0131r.<\/p>\n

\u00c7\u00f6z\u00fcm Y\u00f6ntemi: \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi<\/h2>\n

Bu problemi \u00e7\u00f6zmek i\u00e7in en verimli y\u00f6ntemlerden biri, iki i\u015faret\u00e7i tekni\u011fini kullanmakt\u0131r. \u00d6ncelikle, diziyi s\u0131ralayarak ba\u015fl\u0131yoruz. Ard\u0131ndan, dizideki her bir eleman\u0131 ele al\u0131p, hedef toplam\u0131 (0) elde etmek i\u00e7in kalan iki say\u0131y\u0131 bulmak \u00fczere iki i\u015faret\u00e7i kullan\u0131yoruz. Bu i\u015faret\u00e7iler, s\u0131ral\u0131 dizinin kalan k\u0131sm\u0131nda ilerleyerek, toplam\u0131 s\u0131f\u0131ra e\u015fit olan kombinasyonlar\u0131 arar. Bu yakla\u015f\u0131m, \u00e7\u00f6z\u00fcm\u00fc O(n^2) zaman karma\u015f\u0131kl\u0131\u011f\u0131 ile ger\u00e7ekle\u015ftirir, bu da olduk\u00e7a verimlidir. Ayr\u0131ca, benzersiz kombinasyonlar\u0131 bulmak i\u00e7in ek kontroller eklememiz gerekir.<\/p>\n

JavaScript Kodu<\/h2>\n
\n\/**\n * @param {number[]} nums\n * @return {number[][]}\n *\/\nvar threeSum = function(nums) {\n    nums.sort((a, b) => a - b); \/\/ Dizinin s\u0131ralanmas\u0131\n    const result = [];\n    for (let i = 0; i < nums.length - 2; i++) {\n        \/\/ Tekrar eden say\u0131lar\u0131 atlamak i\u00e7in\n        if (i > 0 && nums[i] === nums[i - 1]) continue; \n        let left = i + 1;\n        let right = nums.length - 1;\n        while (left < right) {\n            const sum = nums[i] + nums[left] + nums[right];\n            if (sum === 0) {\n                result.push([nums[i], nums[left], nums[right]]);\n                \/\/ Tekrar eden say\u0131lar\u0131 atlamak i\u00e7in\n                while (left < right && nums[left] === nums[left + 1]) left++;\n                while (left < right && nums[right] === nums[right - 1]) right--;\n                left++;\n                right--;\n            } else if (sum < 0) {\n                left++;\n            } else {\n                right--;\n            }\n        }\n    }\n    return result;\n};\n<\/code><\/pre>\n
Kod A\u00e7\u0131klamas\u0131<\/h2>\n
Kod \u00f6ncelikle diziyi s\u0131ralar. Sonras\u0131nda, d\u0131\u015f d\u00f6ng\u00fc, dizinin her bir eleman\u0131n\u0131 teker teker inceler.  \u0130\u00e7 d\u00f6ng\u00fc ise, iki i\u015faret\u00e7i (left<\/code> ve right<\/code>) kullanarak, toplam\u0131 s\u0131f\u0131ra e\u015fit olan kombinasyonlar\u0131 arar.  E\u011fer toplam s\u0131f\u0131rdan k\u00fc\u00e7\u00fckse, left<\/code> i\u015faret\u00e7isi sa\u011fa kayd\u0131r\u0131l\u0131r; toplam s\u0131f\u0131rdan b\u00fcy\u00fckse, right<\/code> i\u015faret\u00e7isi sola kayd\u0131r\u0131l\u0131r.  Benzersiz kombinasyonlar bulmak i\u00e7in gerekli kontroller de eklenmi\u015ftir.  Sonu\u00e7 olarak, t\u00fcm benzersiz \u00fc\u00e7l\u00fc kombinasyonlar d\u00f6nd\u00fcr\u00fcl\u00fcr.<\/p>\n
Karma\u015f\u0131kl\u0131k Analizi<\/h2>\n
Bu \u00e7\u00f6z\u00fcm\u00fcn zaman karma\u015f\u0131kl\u0131\u011f\u0131 O(n^2) ve yer karma\u015f\u0131kl\u0131\u011f\u0131 O(1)’dir (yerinde s\u0131ralama varsay\u0131m\u0131yla).  S\u0131ralama i\u015flemi O(n log n) zaman al\u0131rken, iki i\u015faret\u00e7i d\u00f6ng\u00fcs\u00fc O(n^2) zaman al\u0131r.  Ancak, s\u0131ralama i\u015flemi genel zaman karma\u015f\u0131kl\u0131\u011f\u0131na bask\u0131n gelmez, \u00e7\u00fcnk\u00fc  iki i\u015faret\u00e7i tekni\u011fi O(n^2)  zaman almaktad\u0131r.<\/p>\n
Sonu\u00e7<\/h2>\n
Bu makalede, LeetCode 3Sum problemini \u00e7\u00f6zmek i\u00e7in verimli bir JavaScript \u00e7\u00f6z\u00fcm\u00fc sunduk.  \u0130ki i\u015faret\u00e7i tekni\u011fi kullanarak,  problemi O(n^2) zaman karma\u015f\u0131kl\u0131\u011f\u0131 ile \u00e7\u00f6zebildik.  Umar\u0131m bu makale,  algoritma ve veri yap\u0131lar\u0131 becerilerinizi geli\u015ftirmenize yard\u0131mc\u0131 olur. Daha fazla algoritma \u00e7\u00f6z\u00fcm\u00fc ve yaz\u0131l\u0131m geli\u015ftirme ipu\u00e7lar\u0131 i\u00e7in fatihsoysal.com<\/a> adresini ziyaret edebilirsiniz.<\/p>\n
Faydal\u0131 Kaynaklar:<\/strong><\/p>\n