[LeetCode] 18. 4Sum

Given an array S of n integers, are there elements abc, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note: The solution set must not contain duplicate quadruplets.
For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

Thought process:
Add one loop to "15. 3Sum".

Solution:
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public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        List<List<Integer>> quadruplets = new ArrayList<>();
        
        for (int i = 0; i < nums.length - 3; i++) {
            while (i > 0 && i < nums.length - 3 && nums[i] == nums[i - 1]) {
                i++;
            }
            
            for (int j = i + 1; j < nums.length - 2; j++) {
                while (j > i + 1 && j < nums.length - 2 && nums[j] == nums[j - 1]) {
                    j++;
                }
                if (j >= nums.length) {
                    break;
                }
                
                int start = j + 1;
                int end = nums.length - 1;
                while (start < end) {
                    int fourSum = nums[i] + nums[j] + nums[start] + nums[end];
                    if (fourSum < target) {
                        start++;
                    } else if (fourSum > target) {
                        end--;
                    } else {
                        List<Integer> quadruplet = new ArrayList<>();
                        quadruplet.addAll(Arrays.asList(nums[i], nums[j], nums[start], nums[end]));
                        quadruplets.add(quadruplet);
                        
                        start++;
                        while (start < nums.length && nums[start] == nums[start - 1]) {
                            start++;
                        }
                        end--;
                        while (end >= 0 && nums[end] == nums[end + 1]) {
                            end--;
                        }
                    }
                }
            }
        }
        
        return quadruplets;
    }
}

Time complexity:
Sorting takes O(nlogn). Loop takes O(n^3). The overall time complexity is O(n^3).

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