LeetCode:4Sum

最新推荐文章于 2025-07-04 02:32:29 发布

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最新推荐文章于 2025-07-04 02:32:29 发布

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原文链接：http://www.cnblogs.com/xiaoying1245970347/p/4553868.html

本文探讨了如何解决四数之和的问题，即在给定的整数数组中找到四个元素，使它们的和等于目标值。通过使用排序和双指针技术，实现了时间复杂度为O(n^3)的解决方案，并进一步优化至O(n^2)，通过利用哈希表缓存两个数的和。

problems:

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)

Solution:依然采用先排序 然后左右夹逼 时间复杂度O(n3)  同2sum

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        
        vector<vector<int>> result;
        if(nums.size()<4) return result;
        sort(nums.begin(),nums.end());
        
        auto last=nums.end();
        
        for(auto a=nums.begin();a<prev(last,3);++a)
           for(auto b=next(a);b<prev(last,2);++b)
           {
               auto c=next(b);
               auto d=prev(last);
               while(c<d)
               {
                   if(*a+*b+*c+*d<target)
                     ++c;
                     else if(*a+*b+*c+*d>target)
                     --d;
                     else{
                         
                         result.push_back({*a,*b,*c,*d});
                         ++c;
                         --d;
                     }
               }
              
           }
           sort(result.begin(),result.end());
           result.erase(unique(result.begin(),result.end()),result.end());
           return result;
        
    }
};

尽可能的降低时间复杂度，我们可以用一个hashmap来缓存两个数的和，这样子时间复杂度可以降低到O(n2)

转载于:https://www.cnblogs.com/xiaoying1245970347/p/4553868.html