3Sum Java

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)

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Check the coding below in detail:

 /*
    Better Solution: Time: O(n^2)
    Key to solve: HashSet, use 2 additional index inside a loop
    Idea: Always sort Array first
    iterator array from beginning, use additional two  index start=i+1 and end=length+1
    Thus, there are 3 Cases:
    1. a==b+c => add into list
    2. a>b+c  => start++
    3. a<b+c  => end--
    */
    public List<List<Integer>> threeSum(int[] num) {
        List<List<Integer>> triplets =new ArrayList<List<Integer>>();
        ArrayList<Integer> sub=new ArrayList<Integer>();
        HashSet<ArrayList<Integer>> set=new HashSet<ArrayList<Integer>>();
        Arrays.sort(num);
        int len=num.length;
        for(int i=0;i<len-2;i++){
            int start=i+1;
            int end=len-1;
            int negate=(-num[i]);
            while(start<end){
                if((num[start]+num[end])==negate){
                    sub.add(num[i]);
                    sub.add(num[start]);
                    sub.add(num[end]);
                    if(!set.contains(sub)){
                        set.add(sub);
                        triplets.add(new ArrayList<Integer>(sub));

                    }
                    //move the indexs
                    start++;
                    end--;
                    sub.clear();
                }
                else if(negate>num[start]+num[end]){
                    start++;
                }else{
                    end--;
                }
            }
        }
        return triplets;
    }