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).

题目：找到三个数，这三个数的和与target最近接，返回这三个数的和；

思路：

先排序，然后左右各一指针，在sum小于目标值的时候左指针向右走，在sum大于目标值的时候右指针向左走。

代码：

#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){//找到与target最接近的三个数之和  并且三者的地址vector<int>a = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };int target = 6;int dist = INT_MAX;sort(a.begin(), a.end());int minsum;vector<vector<int>::const_iterator> index;for (vector<int>::const_iterator it = a.begin(); it != a.end(); it++){vector<int>::const_iterator first = it + 1;vector<int>::const_iterator last = a.end() - 1;while (first < last){int sum = *it + *first + *last;//三个数的和对target进行判断/*index.push_back(it);index.push_back(first);index.push_back(last);*/if (sum>target){if (abs(sum - target) < dist){dist = abs(sum - target);minsum = dist;}last--;}if (sum < target){if (abs(sum - target) < dist){dist = abs(sum - target);minsum = dist;}first++;}else{dist = 0;minsum = target;break;}}}cout << minsum <<"  "<<dist<< endl;system("pause");return 0;}

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {                int sum=INT_MAX;        int minsum=0;        int nsum;        sort(nums.begin(),nums.end());        int i,j,k;        for(i=0;i<nums.size();i++)        {            j=i+1;            k=nums.size()-1;            while(j<k)            {                nsum=nums[i]+nums[j]+nums[k];                              if(abs(nsum-target)<sum)                {                    sum=abs(nsum-target);                    minsum=nsum;                }                if(nsum-target<0)                {                    j++;                }                else if(nsum-target>0)                {                                        k--;                }                else if((nsum-target)==0)                {                   return target;                }                           }                   }         return minsum;    }};