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题目：给一个数组和给定的目标值，要求在数组里找出三个元素，这三个元素的和最接近目标值，当然等于是最好的。

用3sum的方法，把判定条件作些修改。

int twoSum(vector
   
     &num, int start, int target){	if(num.size() < 3 || start >= num.size())		return -target;	int head = start;	int tail = num.size() - 1;	int result = -target;	int erro = 0x7fffffff;	while (head < tail)	{		int sum = num[head] + num[tail];		if(sum == target)			{				result = target;				return target;		    }		else if(sum < target){			do 			{				++head;			} while (head < tail && num[head] == num[head - 1]);		}		else		{			do 			{				--tail;			} while (tail > head && num[tail] == num[tail + 1]);		}		int curerro = abs(sum - target);		if(curerro < erro)		{			result = sum;			erro = curerro;		}	}	return result;}int threeSumClosest(vector
    
      &num, int target) {	const int len = num.size();	if(len < 3)		return 0;	sort(num.begin(), num.end());	int closet = -target;	int erro = 0x7fffffff;	for (int i = 0; i < len - 2; ++i)	{		int thisloop = twoSum(num, i + 1, target - num[i]);		if(thisloop == target - num[i])			return target;		int cur_erro = abs(thisloop + num[i] - target);	    if(cur_erro < erro)		  {			  closet = thisloop + num[i];			  erro = cur_erro;		}		while(i + 1 < len && num[i + 1] == num[i]) 			++i;	}	return closet;}
    
   

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