LeetCode 3Sum Closest

converse the question to :

　　find the minimum of (3Sum-target). We can also put the (-target) to the array~

similar to the 3Sum,only change the find condition a bit! 

Note: assume that each input would have exactly one solution

Time Complexity: O(n^2)

Space Complexity: O(1)

bool compare(const int& a, const int& b)
{
      return (a<b);
}


class Solution {


public:
    int threeSumClosest(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        std::sort(num.begin(), num.end(), compare);
        
        int len = num.size();
        if(len == 1)    return num[0];
        
        int minsum = 0x8fffffff;
        int tmpsum, tmp;
        int pl, pr;
                
        for(int i=0; i<len; i++)
        {
            pl = i+1;
            pr = len-1;
            
            while(pl < pr)
            {
                tmpsum = num[pl]+num[pr]+num[i];
                
                if(tmpsum == target)    return target; 
                
                if( abs(tmpsum-target) < abs(minsum-target) )
                    minsum = tmpsum;
                
                if(tmpsum < target)    pl++;
                else    pr--;    
            }
        }
        
        return minsum;
    }
};

References:

http://blog.csdn.net/li4951/article/details/8693212