Leetcode:3Sum Closest

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：有序数组找两个数和为target的变体，前后两个指针的O（n）变体。

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        //TODO input check
        int ithreeclose=0;
        if(target>=0)
            ithreeclose=INT_MAX;
        else
            ithreeclose=INT_MIN;
        sort(num.begin(),num.end() );
        for(int i=0; i<num.size()-2; ++i)
        {
            int itwoclose=twoSumClosest(num,i+1,target-num[i]);   
            if(abs(itwoclose+num[i]-target) < abs(ithreeclose-target) )
            {
                ithreeclose = itwoclose+num[i]; 
            }
        }
        return ithreeclose;
    }
    int twoSumClosest(vector<int>& num,int ibeg, int target)
    {
        int iclose=0;
        if(target>=0)
            iclose=INT_MAX;
        else
            iclose=INT_MIN;
        int iend=num.size()-1;  
        while(ibeg < iend)
        {
            if(num[ibeg]+num[iend] == target)
                //break;
                return target;
            else if(num[ibeg]+num[iend] < target)
            {
                if(abs(num[ibeg]+num[iend]-target) < abs(iclose-target) )
                    iclose = num[ibeg]+num[iend]; 
                ++ibeg; 
            }
            else            
            {
                if(abs(num[ibeg]+num[iend]-target) < abs(iclose-target) )
                    iclose = num[ibeg]+num[iend]; 
                --iend; 
            }

        }
        return iclose;
    }
};