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). 我的思路： 　　　　（1）首先对给定的数组进行排序，让里面的数值从小到大； 　　　　（2）遍历数组，用一个for循环控制作为外循环，下标从小到大进行遍历，因为需要3个数，所以用外循环的这个下标对应的数组值作为base，然后再 　　　　　　　　初始化一个begin作为base的下一位置，end作为数组末尾，三者相加得到sum，若是value=sum-target（目标值）更小则保存sum的值，并且若value<0 　　　　　　　　则begin++，若value>o则end--，若是value=0，则返回target； 看代码：

class Solution {public:    int threeSumClosest(vector
     
      & nums, int target);     int  findClosest(vector
      
       & nums, int base, int end, int target); private:    int result, finalResult, gap = 99999999, gap1 = 99999999;};int Solution::threeSumClosest(vector
       
        & nums, int target){    //int result,gap=99999999;    sort(nums.begin(), nums.end());    int len = nums.size();    for (int i = 0; i < nums.size() - 2; i++)    {        //cout << "before" << endl;        int res1 = findClosest(nums, i, len - 1, target);        if (res1 < gap1)            gap1 = res1;        finalResult = result;    }    //    cout << "over" << endl;    //    return gap1;    return finalResult;}int Solution::findClosest(vector
        
         & nums, int base, int end, int target){    //cout << "executing" << endl;    int begin = base + 1;    while (begin < end){        int value = nums[base] + nums[begin] + nums[end] - target;        int temp = abs(nums[base] + nums[begin] + nums[end] - target);        if (temp <= gap){            //        int s = nums[base], s2 = nums[begin], s3 = nums[end];            gap = abs(nums[base] + nums[begin] + nums[end] - target);            result = nums[base] + nums[begin] + nums[end];            if (value == 0)                return 0;            else if (value>0)                end--;            else if (value < 0)                begin++;        }        else {             if (value>0) end--;            else begin++;        //    return gap;         }    }    return gap;}