LeetCode:3Sum

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)

Idea:  o(n^2) solution exists. First sort the array, and then from left to right, for each num[i], search the pair that sums up to -num[i] using Two Sum algorithm.


Java Code:

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(num.length<3) return res;
        
        Arrays.sort(num);
        for(int i=0;i<num.length-2;i++){
            if(i==0 || num[i]>num[i-1]){ //avoid duplicate solutions   
                int j=i+1, 
                    k=num.length-1;
         
                while(j<k){ 
                    if(num[j]+num[k]==-num[i]){
                        ArrayList<Integer> temp = new ArrayList<Integer>();
                        temp.add(num[i]);
                        temp.add(num[j]);
                        temp.add(num[k]);
                        res.add(temp);
                        k--;
                        j++;
                        while(k>j && num[k]==num[k+1]) k--;//avoid duplicate solutions 

                        while(j<k && num[j]==num[j-1]) j++;//avoid duplicate solutions 

                    }else if(num[j]+num[k]>-num[i]){
                        k--;
                    }else{
                        j++;
                    }
                }
            }
        }
        return res;
    }
}