3Sum Closest

题目：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).

思想：采用3SUM的方法，设置两个指针，一个指向nums[1]，一个指向nums[n],然后两者不断逼近。

代码：

#include <iostream>#include <vector>#include <algorithm>using namespace std;class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {       if(nums.size()<3)return 0;sort(nums.begin(),nums.end());int sum;int result=nums[0]+nums[1]+nums[nums.size()-1];//求出第一个和for(int i=0;i<nums.size()-2;++i){int j=i+1;int n=nums.size()-1;while(j<n){sum=nums[i]+nums[j]+nums[n];if(abs(sum-target)<abs(result-target))result=sum;else{if(sum>target){n--;}else if(sum==target)return target;else{j++;}}}}return result;    }};int main(){Solution s;int arr[]={0,2,1,-3};vector<int> ivec(arr,arr+4);int  ret=s.threeSumClosest(ivec,1);cout<<ret<<endl;system("pause");return 0;}