leetcode: 4Sum

最新推荐文章于 2025-07-04 02:32:29 发布

原创最新推荐文章于 2025-07-04 02:32:29 发布 · 384 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

编程专栏收录该内容

27 篇文章

订阅专栏

本文介绍了一种通过排序和夹逼法解决给定数组中是否存在四个元素之和等于目标值的问题。算法的时间复杂度为O(n^3)，空间复杂度为O(1)，并详细解释了如何避免重复解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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)

分析：
先排序，然后左右夹逼
T(n)=O(n^3) 空间复杂的：T(n)=O(1)

 class Solution {
 public:
     vector<vector<int>> fourSum(vector<int>& nums, int target) {
         int size = nums.size();
         vector<int> vec;
         vector<vector<int>> ret;
         if (size < 4)
             return ret;
         sort(nums.begin(), nums.end());
         for (int i = 0; i < size-3; i++){
             for (int j = i + 1; j < size-2; j++){
                 int k = j + 1,u=size-1;
                 while (k<u){
                     if (nums[i] + nums[j] + nums[k] + nums[u] == target){
                         ret.push_back({ nums[i], nums[j], nums[k], nums[u] });
                         k++;
                        u--;
                        int tmp = u +1;
                        while (u >= 0 && tmp >= 0 && nums[tmp] == nums[u]){
                                tmp--;
                                u--;
                        }

                     }
                     else if (nums[i] + nums[j] + nums[k] + nums[u] > target)
                         u--;
                     else
                         k++;
                 }  
                 int tmp = j + 1;
                 while (j < size && tmp < size && nums[tmp] == nums[j]){
                     tmp++;
                     j++;
                 }          
             }
             int tmp = i + 1;
             while (i < size && tmp < size && nums[tmp] == nums[i]){
                 tmp++;
                 i++;
             }

         }
         return ret;
     }
 };