【LeetCode】3Sum Closest

题目描述：

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).

这题跟前面的 3Sum类似，但是因为求的不是指定值，没法用map了。

用三重遍历试了下，不出意外的TLE。

O（n^2）解法能想到的只有这种：

对数组排序后，首先确定一个值，另外两个值按如下规则从两边向中间挤：若和小于target，则left++，else则right--。

代码如下：

class Solution {
public:
	int threeSumClosest(vector<int> &num, int target) {
		int res = num[0] + num[1] + num[2];
		int N = num.size();
		sort(num.begin(), num.end());
		for (int i = 0; i < N; i++){
			int j = i + 1, k = N - 1;
			int dec(INT_MAX);
			while (j < k){
				int sum = num[i] + num[j] + num[k];
				if (abs(sum - target)>dec)
					break;
				dec = abs(sum - target);
				res = dec < abs(res - target) ? sum : res;
				if (sum>target)
					k--;
				else
					j++;
			}
		}
		return res;
	}
};