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).

该题目与类似，但并不是列出所有和为0的三元组，而是找出与target最相近的sum，同样是利用三个指针i，p和q，具体请参考。只不过在遍历时记录最小误差minerror，最后返回target+minerror即可。

代码如下：

1 class Solution { 2 public: 3     int threeSumClosest(vector
     
       &num, int target)  4     { 5         sort(num.begin(), num.end()); 6         int minerror = num[0]+num[1]+num[2] - target; 7         for( int i = 0 ; i < num.size() ; i++ ) 8         { 9             if( i > 0 && num[i] == num[i-1] )10             {11                 continue;12             }13             int p = i+1;14             int q = num.size() - 1;15             while( p < q )16             {17                 if( p > i +1 && num[p] == num[p-1] )18                 {19                     p++;20                     continue;21                 }22 23                 if( q < num.size() -1 &&  num[q] == num[q+1] )24                 {25                     q--;26                     continue;27                 }28 29                 int sum = num[i] + num[p] + num[q]; 30 31                 if( abs(sum - target) < abs(minerror) )32                 {33                     minerror = sum - target;34                 }35 36                 if( sum - target  == 0)37                 {38                     return target;39                 }40 41                 if( sum - target > 0 )42                 {43                     q--;44                 }45 46                 if( sum - target < 0 )47                 {48                     p++;49                 }50             }51         }52             return minerror + target;53     }54 };