Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

  • 0 <= a, b, c, d < n
  • a, b, c, and d are distinct.
  • nums[a] + nums[b] + nums[c] + nums[d] == target

You may return the answer in any order.

Example 1:

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

Example 2:

Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]

Constraints:

  • 1 <= nums.length <= 200
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109

四数之和。

给定一个包含 n 个整数的数组 nums 和一个目标值 target，判断 nums 中是否存在四个元素 a，b，c 和 d ，使得 a + b + c + d 的值与 target 相等？找出所有满足条件且不重复的四元组。

注意：答案中不可以包含重复的四元组。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/4sum
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

依然是双指针逼近的思路做，注意以下几点

  • 需要对input排序
  • 需要四个指针，i, j, low, high
  • 每个指针都需要跳过重复元素
  • i最多到nums.length - 3

Java实现

 1 class Solution {
 2     public List<List<Integer>> fourSum(int[] nums, int target) {
 3         List<List<Integer>> res = new ArrayList<>();
 4         if (nums.length < 4) {
 5             return res;
 6         }
 7         Arrays.sort(nums);
 8         for (int i = 0; i < nums.length - 3; i++) {
 9             if (i > 0 && nums[i] == nums[i - 1]) {
10                 continue;
11             }
12             for (int j = i + 1; j < nums.length - 2; j++) {
13                 if (j > i + 1 && nums[j] == nums[j - 1]) {
14                     continue;
15                 }
16                 int low = j + 1;
17                 int high = nums.length - 1;
18                 while (low < high) {
19                     int sum = nums[i] + nums[j] + nums[low] + nums[high];
20                     if (sum == target) {
21                         res.add(Arrays.asList(nums[i], nums[j], nums[low], nums[high]));
22                         while (low < high && nums[low] == nums[low + 1]) {
23                             low++;
24                         }
25                         while (low < high && nums[high] == nums[high - 1]) {
26                             high--;
27                         }
28                         low++;
29                         high--;
30                     } else if (sum < target) {
31                         low++;
32                     } else {
33                         high--;
34                     }
35                 }
36             }
37         }
38         return res;
39     }
40 }

 

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @param {number} target
 4  * @return {number[][]}
 5  */
 6 var fourSum = function(nums, target) {
 7     nums = nums.sort((a, b) => a - b);
 8     const res = [];
 9     let low, high, sum;
10     // corner case
11     if (nums.length < 4) return res;
12 
13     // normal case
14     for (let i = 0; i < nums.length - 3; i++) {
15         if (i > 0 && nums[i] === nums[i - 1]) continue;
16         for (let j = i + 1; j < nums.length - 2; j++) {
17             if (j > i + 1 && nums[j] === nums[j - 1]) continue;
18             low = j + 1;
19             high = nums.length - 1;
20             while (low < high) {
21                 sum = nums[i] + nums[j] + nums[low] + nums[high];
22                 if (sum === target) {
23                     res.push([nums[i], nums[j], nums[low], nums[high]]);
24                     while (low < high && nums[low] === nums[low + 1]) low++;
25                     while (low < high && nums[high] === nums[high - 1]) high--;
26                     low++;
27                     high--;
28                 } else if (sum < target) {
29                     low++;
30                 } else {
31                     high--;
32                 }
33             }
34         }
35     }
36     return res;
37 };

 

LeetCode 题目总结