3Sum Closest(离目标值最近的三数之和)

    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).
    解析：(1)对数组进行排序。(2)遍历数组中的每个元素。设置两个指针，start指向该元素后一个位置，last指向数组的末尾。(3)求出这三个数之和与目标值得差值，如果当前差值小于上一次的差值，则把当前差值更新为最小的差值。同时更新最后结果为三数之和。最后返回结果即可。

int threeSumClosest(vector& nums, int target)
{
	//最近距离
	int closestNum = INT_MAX;
	//三个数之和
	int res = 0;
	//当数组长度小于三时，返回最大整数
	if (nums.size() < 3)
		return closestNum;
	//对数组进行排序
	sort(nums.begin(), nums.end());
	//遍历数组
	for (int i = 0; i < nums.size() - 2; ++i)
	{
		int start = i + 1;
		int last = nums.size() - 1;
		while (start < last)
		{
			//当前三个数之和与目标值得差值
			int dis = nums[i] + nums[start] + nums[last] - target;
			//如果小于上一次的距离值，则更新最小距离和三个数之和
			if (abs(dis) < closestNum)
			{
				closestNum = abs(dis);
				res = nums[i] + nums[start] + nums[last];
			}
			if (dis < 0)
			{
				start++;
			}
			else if (dis > 0)
			{
				last--;
			}
			else
			{
				return res;
			}
		}
	}
	return res;
}