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).

解题思路:

    与Three Sum的做法类似，首先枚举中间数字的位置，然后找寻和离target减去中间数字最近的点即可(此处需用绝对值).时间复杂度为O(N^2).

解题代码:

    
class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) 
    {
        sort(num.begin(),num.end());
        unsigned n = num.size() ;
        long long res = abs(num[0] + num[1] + num[2] - target) , sum = num[0] + num[1] + num[2] ;
        for(unsigned i=1;i < n - 1 ;++i)
        {
            long long sum1 = target - num[i] ;
            unsigned j = 0 , k = n - 1 ;
            while( j < i && k > i )
            {
                long long tmp = num[j] + num[k] ;
                if(res > min(res,abs(sum1-tmp)))
                {
                    res = min(res,abs(sum1-tmp));
                    sum = num[i] + tmp;
                }                    
                if(sum1 == tmp)
                    return sum;
                tmp > sum1 ? --k : ++j ;
            }
        }
        return sum;
    }
};