Three sum and result total =0

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.

 

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
 

Constraints:

3 <= nums.length <= 3000
-105 <= nums[i] <= 105

solution:

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums);

        for (int i = 0; i < nums.length - 2 && nums[i] <= 0; i++) {
            if (i != 0 && nums[i] == nums[i - 1])
                continue;
            twoSum(-nums[i], nums, i + 1, result);
        }

        return result;
    }

    private void twoSum(int target, int[] nums, int startingIndex, List<List<Integer>> result) {
        int l = startingIndex, r = nums.length - 1;

        while (l < r) {
            if (nums[l] + nums[r] > target) {
                r--;
                continue;
            }

            if (nums[l] + nums[r] < target) {
                l++;
                continue;
            }
            result.add(Arrays.asList(-target, nums[l], nums[r]));
            l++;
            r--;

            while (r > l && nums[r] == nums[r + 1])
                r--;
        }
    }
}