15 3Sum – Medium

Problem:

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)

Thoughts:

Key idea here is to sort the array first. Then start from left to pick the first element to be a, then find if there are two other elements b, c make a + b + c = 0. Another thing to keep in mind is that how to avoid duplicate answers.

Solutions:

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new LinkedList<List<Integer>>();
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if (i == 0 || nums[i] != nums[i-1]) {
                int start = i + 1, end = nums.length - 1;
                while (start < end) {
                    int sum = nums[i] + nums[start] + nums[end];
                    if (sum == 0){
                        List<Integer> tmp = new LinkedList<Integer>();
                        tmp.add(nums[i]);
                        tmp.add(nums[start]);
                        tmp.add(nums[end]);
                        result.add(tmp);
                        int startVal = nums[start];
                        int endVal = nums[end];
                        while (start < end && startVal == nums[start]) {
                            start ++;
                        }
                        while (end > start && endVal == nums[end]) {
                            end --;
                        }
                    }
                    else if (sum < 0) {
                        start ++;
                    }
                    else {
                        end --;
                    }
                }
            }
        }
        return result;
    }
 
}