[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)

Solution: 
Sorting helps us to deal with both constraints. Keep 3 pointers. We move the 1st pointer from left to right, and each time solve a 2SUM problem using the other 2 pointers in the subarray with index ranging from i + 1 to n.
Complexity: O(n^2). 
Since 2SUM can be solved in O(n) time after sorting, or in O(n) time using HashMap.

public class Solution {
    private ArrayList<Integer> oneTriple(int a, int b, int c) {
        ArrayList<Integer> triple = new ArrayList<Integer>();
        triple.add(a);
        triple.add(b);
        triple.add(c);
        return triple;
    }
    
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        ArrayList<List<Integer>> list 

            = new ArrayList<List<Integer>>();
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int j = i + 1;
            int k = nums.length - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum == 0) {
                    list.add(oneTriple(nums[i], nums[j], nums[k]));
                    j++;
                    k--;
                    while (j < k && nums[j] == nums[j - 1]) {
                        j++;
                    }
                    while (j < k && nums[k] == nums[k + 1]) {
                        k--;
                    }
                } else if (sum > 0) {
                    k--;
                } else {
                    j++;
                }
            }
        }
        return list;
    }
}