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] ]
class Solution { public List<List<Integer>> threeSum(int[] nums) { Arrays.sort(nums); List<List<Integer>> res = new LinkedList<>(); for (int i = 0; i < nums.length - 2; i++) { if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) { int lo = i + 1; int hi = nums.length - 1; int sum = 0 - nums[i]; while (lo < hi) { if (nums[lo] + nums[hi] == sum) { res.add(Arrays.asList(nums[i], nums[lo], nums[hi])); while (lo < hi && nums[lo] == nums[lo + 1]) //略过重复的元素 lo++; while (lo < hi && nums[hi] == nums[hi - 1]) hi--; lo++; hi--; } else if (nums[lo] + nums[hi] < sum) { while (lo < hi && nums[lo] == nums[lo + 1]) lo++; lo++; } else { while (lo < hi && nums[hi] == nums[hi - 1]) hi--; hi--; } } } } return res; } } 标签：15,nums,3Sum,lo,int,while,hi,&& From： https://www.cnblogs.com/MarkLeeBYR/p/16850601.html