4Sum leetcode

Given an array S of n integers, are there elements a, b, c, 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:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• 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)

与3sum相同 但是多了一层循环 时间复杂O(n^3)

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (nums == null || nums.length == 0) {
            return res;
        }
        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 3; i++) {
            if (i > 0 && nums[i] == nums[i-1]) {//如果i重复取 结果相同
                continue;
            }
            for (int j = i + 1; j < nums.length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                int left = j + 1;
                int right = nums.length - 1;
                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        List<Integer> tem = new ArrayList<Integer>();
                        tem.add(nums[i]);
                        tem.add(nums[j]);
                        tem.add(nums[left]);
                        tem.add(nums[right]);
                        res.add(tem);
                        left++;
                        right--;
                        while (left < right && nums[left] == nums[left - 1]) {
                            left++;
                        }
                        while (left < right && nums[right] == nums[right + 1]) {
                            right --;
                        }
                    } else if (sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }
        return res;
    }
}