Given an array nums of n integers, are there elements a, b, c in nums 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.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]] 有空没空也得记得刷道题啊，不然乍一敲代码手生的很啊。

1 class Solution { 2     public List
     
      
       > threeSum(int[] nums) { 3         List
       
        
         > list = new ArrayList<>();//存放List
         
          数组元素 4         Arrays.sort(nums); 5         int length = nums.length; 6         for (int i = 0; i < length-2; i++) { 7             if (i != 0 && nums[i] == nums[i-1]) 8                 continue; 9             for (int j = i+1; j < length-1; j++) {10                 if (nums[i] + nums[j] + nums[length-1] < 0)11                     continue;12                 int th = nums[i] + nums[j];13                 if(th > 0)//确保nums[i]+nums[j] < 014                     continue;15 16                 for (int i1 = nums.length - 1; i1 > j; i1--) {17                     if (nums[i1] + th == 0) {18                         if (list.size() > 0) {19                             List
          
            l = list.get(list.size()-1);20 if (l.get(0) == nums[i] && l.get(1) == nums[j] && l.get(2) == nums[i1]) {21 break;22 }23 }24 ArrayList
           
             listElem = new ArrayList<>();25 listElem.add(nums[i]);26 listElem.add(nums[j]);27 listElem.add(nums[i1]);28 list.add(listElem);29 break;30 }31 else if (nums[i1] + th < 0)32 break;33 }34 }35 }36 37 return list;38 }39 }