leetcode -- 3Sum

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)

 [Some tricks]
1. Line 18 and Line 23.
    filter the duplicate during two-pointer scan. For example [-2, 0, 0, 2,2], the expected output should be [-2,0,2]. If no filter here, the output will be duplicate as [-2,0,2] and [-2,0,2]
2. Line 31-33
   filter the duplicate for outside iteration. For example [-2, -2, -2, 0,2].

public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        Arrays.sort(num);
        int length = num.length;
        for(int i = 0; i < length; i++){
            int start = i + 1;
            int end = length - 1;
            int target = 0 - num[i];
            while(start < end){
                if(target == num[start] + num[end]){
                    ArrayList<Integer> list = new ArrayList<Integer>();
                    list.add(num[i]);
                    list.add(num[start]);
                    list.add(num[end]);
                    result.add(list);
                    start ++;
                    end --;
                    while(num[start] == num[start - 1] && start < end){
19                         start ++;
20                     }
21                     while(num[end] == num[end + 1] && start < end){
22                         end --;
23                     }
                } else if(target > num[start] + num[end]){
                    start ++;
                } else {
                    end --;
                }
            }
            
            while ((i < length - 1) && num[i] == num[i + 1]) {
32                 i++;
33             } 
        }
        return result;
    }