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).

思路：

类似
http://blog.csdn.net/lanxu_yy/article/details/11896109。

代码：

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(num.size() < 3)
        {
            return INT_MIN;
        }
        
        int min = INT_MAX;
        sort(num.begin(), num.end());
        
        for(int i = 0; i < num.size()-2; i++)
        {
            if(i > 0 && num[i] == num[i-1])
            {
                continue;
            }
            for(int j = i+1; j < num.size()-1; j++)
            {
                if(j > i+1 && num[j] == num[j-1])
                {
                    continue;
                }
                int left = j+1;
                int right = num.size()-1;
                while(left<=right)
                {
                    int mid = left + ((right-left)>>1);
                    if(num[i] + num[j] + num[mid] == target)
                    {
                        return target;
                    }
                    else if(num[i] + num[j] + num[mid] > target)
                    {
                        int m = num[i] + num[j] + num[mid];
                        if(min == INT_MAX)
                        {
                            min = m;
                        }
                        else
                        {
                            min = abs(m-target)<abs(min-target)?m:min;
                        }
                        right = mid - 1;
                    }
                    else
                    {
                        int m = num[i] + num[j] + num[mid];
                        if(min == INT_MAX)
                        {
                            min = m;
                        }
                        else
                        {
                            min = abs(m-target)<abs(min-target)?m:min;
                        }
                        left = mid + 1;
                    }
                };
            }
        }
        
        return min;
    }
};