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).

與3Sum極爲類似。只是處理時判斷哪個更靠近。靠近即距離，需絕對值！！

 int threeSumClosest(vector<int> &num, int target)
    {
    	if(num.size() < 3) return INT_MAX;
    
    	int min_gap = INT_MAX;
    	sort(num.begin(),num.end());
    
    	for(int i = 0;i <= num.size()-3;i++)
    	{
    		if(i > 0 && num[i] == num[i-1])continue;//前面計算過直接跳過
    
    		int left = i+1;
    		int right = num.size() -1 ;
    
    		while(left < right)
    		{
    			int sum = (num[i] + num[left] + num[right] );
    			min_gap = abs(sum - target) < abs(min_gap)?sum - target:min_gap ;//關鍵！比較最小距離
    
    			if(sum == target) 
    			{
    				return target;
    			}
    			else if(sum > target )
    			{
    				
    				right--;
    			}
    			else
    			{
    				left++;
    			}
    		}
    	}
    	return min_gap + target;
    }