首页 > 代码库 > 3Sum Closest
3Sum Closest
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).固定一个数，然后从剩余的元素中选择两个。
C++实现代码：
#include<iostream>#include<algorithm>#include<vector>#include<climits>#include<cmath>using namespace std;class Solution{public: int threeSumClosest(vector<int> &num, int target) { if(num.empty()) return 0; sort(num.begin(),num.end()); int minSum=0; int close=INT_MAX; int sum; int n=num.size(); int i; for(i=0; i<n-2; i++) { int left=i+1; int right=n-1; while(left<right) { if(num[i]+num[left]+num[right]<target) { sum=num[i]+num[left]+num[right]; cout<<sum<<endl; if(target-sum<close) { close=target-sum; minSum=sum; } left++; } else if(num[i]+num[left]+num[right]>target) { sum=num[i]+num[left]+num[right]; cout<<sum<<endl; if(sum-target<close) { close=sum-target; minSum=sum; } right--; } else if(num[i]+num[left]+num[right]==target) { close=0; //cout<<"i "<<i<<" left "<<left<<" right "<<right<<endl; //cout<<num[i]<<" "<<num[left]<<" "<<num[right]<<endl; minSum=num[i]+num[left]+num[right]; return minSum; } } } return minSum; }};int main(){ vector<int> vec= {1,2,3,4}; Solution s; int result=s.threeSumClosest(vec,1); cout<<result<<endl;}
运行结果：
3Sum Closest
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。