【Leetcode】3Sum (Sum)

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},    A solution set is:    (-1, 0, 1)    (-1, -1, 2)

这道题首先对num数组排序，然后求0~i-1里面加和等于-num[i]的数

因为它要求找唯一的组合，所以需要避免重复

代码来自Code_Ganker

public ArrayList<ArrayList<Integer>> threeSum(int[] num) {ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();if (num == null || num.length <= 2)return result;Arrays.sort(num);for (int i = num.length - 1; i >= 2; i--) {if (i < num.length - 1 && num[i] == num[i + 1])continue;ArrayList<ArrayList<Integer>> twoRes = twoSum(num, i - 1, -num[i]);for (int j = 0; j < twoRes.size(); j++)twoRes.get(j).add(num[i]);result.addAll(twoRes);}return result;}public ArrayList<ArrayList<Integer>> twoSum(int[] num, int end, int target) {ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();int left = 0;int right = end;while (left < right) {if (num[left] + num[right] == target) {ArrayList<Integer> temp = new ArrayList<Integer>();temp.add(num[left]);temp.add(num[right]);result.add(temp);left++;right--;while (left < right && num[left] == num[left - 1])left++;while (left < right && num[right] == num[right + 1])right--;} else if (num[left] + num[right] > target)right--;elseleft++;}return result;}