3 Sum 解答

Question

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

First, we sort the array and then implement 2 pointers to get 2 Sum. Time complexity is O(n²)

Notice here, we need to consider duplicates in result.

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        if (nums == null || nums.length < 1)
            return null;
        Arrays.sort(nums);
        int length = nums.length;
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        for (int i = 0; i <= length - 3; i++) {
            // Remove duplicates
            if (i > 0 && nums[i] == nums[i - 1])
                continue;
            int target = 0 - nums[i];
            int l = i + 1, r = length - 1;
            while (l < r) {
                if (nums[l] + nums[r] == target) {
                    result.add(Arrays.asList(nums[i], nums[l], nums[r]));
                    while (l < r && nums[l] == nums[l+1]) l++;
                    while (l < r && nums[r] == nums[r-1]) r--;
                    l++;
                    r--;
                } else if (nums[l] + nums[r] < target) {
                    l++;
                } else {
                    r--;
                }
            }
        }
        return result;
    }
}