LeetCode 15. 3Sum

Question

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

https://leetcode.com/problems/3sum/

• Solution1

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        ArrayList<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length <= 2) return res;
        int n = nums.length;
        int i = 0;
        Arrays.sort(nums);
        while (i < n - 2) {
            int base = nums[i];
            int left = i + 1;
            int right = n - 1;
            while (left < right) {
                int sum = base + nums[left] + nums[right];
                if (sum == 0) {
                    List<Integer> list = new LinkedList<>();
                    list.add(base);
                    list.add(nums[left]);
                    list.add(nums[right]);
                    res.add(list);
                    left = moveRight(nums, left + 1);
                    right = moveLeft(nums, right - 1);
                } else if (sum > 0) {
                    right = moveLeft(nums, right - 1);
                } else {
                    left = moveRight(nums, left + 1);
                }
            }
            i = moveRight(nums, i + 1);
        }
        return res;
    }
    private int moveLeft(int[] nums, int right) {
        while (right == nums.length - 1 || (right >= 0 && nums[right] == nums[right + 1])) {
            right--;
        }
        return right;
    }
     private int moveRight(int[] nums, int left) {
        while (left == 0 || (left < nums.length && nums[left] == nums[left - 1])) {
            left++;
        }
        return left;
    }
}

Complexity:

Time complexity: O( n^2)

Space complexity: O(1)