18. 四数之和 4Sum

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c+ d = target? Find all unique quadruplets in the array which gives the sum of target.

Notice that the solution set must not contain duplicate quadruplets.

Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

方法类似于三数相加

也是双指针，有些情况可以优化，比如最前四个相加大于target可以退出，最后四个相加小于target可以退出。再就是去掉重复的。

public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> ans = new ArrayList<>();
        int n = nums.length;
        if (n < 4) {
            return ans;
        }
        Arrays.sort(nums);
        for (int first = 0; first < n - 3; first++) {
            if (nums[first] + nums[first + 1] + nums[first + 2] + nums[first + 3] > target)
                break;
            if (nums[first] + nums[n - 3] + nums[n - 2] + nums[n - 1] < target)
                continue;
            if (first > 0 && nums[first] == nums[first - 1]) {
                continue;
            }
            for (int second = first + 1; second < n - 2; second++) {
                if (nums[first] + nums[second] + nums[second + 1] + nums[second + 2] > target)
                    break;
                if (nums[first] + nums[second] + nums[n - 2] + nums[n - 1] < target)
                    continue;
                if (second > first + 1 && nums[second] == nums[second - 1]) {
                    continue;
                }
                int forth = n - 1;
                for (int third = second + 1; third < forth; ) {
                    int sum = nums[first] + nums[second] + nums[third] + nums[forth];
                    if (sum == target) {
                        List<Integer> ans_one = new ArrayList<>();
                        ans_one.add(nums[first]);
                        ans_one.add(nums[second]);
                        ans_one.add(nums[third]);
                        ans_one.add(nums[forth]);
                        ans.add(ans_one);
                        while (third < forth && nums[third] == nums[third + 1]) {
                            third++;
                        }
                        third++;
                        while (third < forth && nums[forth] == nums[forth - 1]) {
                            forth--;
                        }
                        forth--;


                    } else if (sum < target) {
                        third++;
                    } else {
                        forth--;
                    }
                }
            }

        }
        return ans;
    }