描述：
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).
思路分析：
这道题让我们求最接近给定值的三数之和，是在之前那道 3Sum 的基础上又增加了些许难度，那么这道题让返回这个最接近于给定值的值，即要保证当前三数和跟给定值之间的差的绝对值最小，所以需要定义一个变量 diff 用来记录差的绝对值，然后还是要先将数组排个序，然后开始遍历数组，思路跟那道三数之和很相似，都是先确定一个数，然后用两个指针 left 和 right 来滑动寻找另外两个数，每确定两个数，求出此三数之和，然后算和给定值的差的绝对值存在 newDiff 中，然后和 diff 比较并更新 diff 和结果 closest 即可。
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>/***@description:Given an array nums of n integers and an integer target, find three integers in nums such that* the sum is closest to target. Return the sum of the three integers.* You may assume that each input would have exactly one solution.Example:Given array nums = [-1, 2, 1, -4], and target = 1.The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).*@author:zjl*@date:2020-1-12*/using namespace std;
/*采用的是夹逼思想*/
class solution{
public:int threesumclosest(vector<int>& nums,int target){int closest=nums[0]+nums[1]+nums[2]; //初始化int diff=abs(closest-target);sort(nums.begin(),nums.end());for(int i=0;i<nums.size();++i){int left=i+1; //left pointerint right=nums.size()-1;//right pointerwhile(left<right){int sum=nums[i]+nums[left]+nums[right];int newdiff=abs(sum-target);if(newdiff<diff){diff=newdiff;closest=sum;}if(sum<target) ++left;else --right;}}return closest;}
};
void display(const int &data)
{cout<<data<<' ';
}
int main(int argc, char *argv[])
{int a[]={-1,2,1,-4};vector<int> vec(a,a+sizeof(a)/sizeof(int));for_each(vec.begin(),vec.end(),display);cout<<endl;solution sl;int result=sl.threesumclosest(vec,1);cout<<result<<endl;return 0;
}