15.3Sum

一,题目

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: 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]]

二,思路

1,先排序
2,固定第一个值的位置,然后从下一所以位置(left) 和最后的索引位置( right )two pointer进行判断
如果固定的位置的值和left,right的和 sum为0,是一个解,,大于零right--, 小于0 left++
在left和right移动过程中,注意相同元素的跳过(skip)即可去除重复的值

三,代码

public  List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> results = new ArrayList<List<Integer>>();if (nums == null || nums.length < 3) {return results;}Arrays.sort(nums);int len = nums.length;for (int i = 0; i < len - 2; i++) {if (i != 0 && nums[i] == nums[i - 1]) {continue;}int left = i + 1;int right = len - 1;int sum = 0;while (left < right) {sum = nums[left] + nums[right] + nums[i];if (sum == 0) {List<Integer> result = new ArrayList<Integer>();result.add(nums[i]);result.add(nums[left]);result.add(nums[right]);results.add(result);left++;right--;while (left < right && nums[left] == nums[left - 1]) {left++;}while (left < right && nums[right] == nums[right + 1]) {right--;}} else if (sum < 0) {left++;} else {right--;}}}return results;}