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).

先对数组进行排序，然后从第一个遍历起，然后设两个指针，在选一个数后，在这个数的后半段开始开始中间收缩，if sum > target则右指针往左移， if sum < target则左指针往右移。

排序O(nlogn) + 查找O(n^2) = O(n^2)

排序O(nlogn) + 查找O(n^2) = O(n^2)

Runtime: 16 ms

//leetcode16.  3Sum Closest
class Solution {
public:
	int threeSumClosest(vector<int> &num, int target) 
	{
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		sort(num.begin(),num.end());
		int ret = 0;
		bool first = true;
		for (int i = 0; i < num.size(); i++)
		{
			int j = i + 1;
			int k = num.size()-1;
			while (j < k)
			{
				int sum = num[i] + num[j] + num[k];
				if (first)
				{
					ret = sum;
					first = false;
				}
				else
				{
					if (abs(sum - target) < abs(ret - target))
						ret = sum;
				}

				if (ret == target)
					return ret;
				if (sum>target)
					k--;
				else
					j++;
			}
		}
		return ret;
	}
};

leetcode.16----------3Sum Closest