16. 3Sum Closest（找出和最接近给定值的三个数）

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).

题目大意：给定一个长度为n的整型数组和一个整数target，在数组中找出三个数，它们的和最接近target。可以认为对于每个测试用例，解决方案都是唯一的。

解题思路：类似《15. 3Sum（求数组中和为0的3个数）》，对于要求三个数的和，可以先对数组从小到大排序，然后锁定一个数nums[i]，根据双指针的思路从数组两端到中间遍历，找到两个数nums[left]和nums[right]，使得nums[left]+nums[right]+nums[i]的和最接近target。遍历过程中使用两个整数diff和res记录当前获得的最优解，其中res表示到目前为止找到的最接近target的三个数的和，diff表示target与res的差的绝对值。当遍历到res==target，已经是最佳方案了，直接返回。

解题代码：（22ms，beats 65.39%）

class Solution {    public int threeSumClosest(int[] nums, int target) {        int diff = Integer.MAX_VALUE;int res = 0;int len = nums.length;Arrays.sort(nums);for (int i = 0; i < len; i++) {int left = i + 1;int right = len - 1;while (left < right) {int sum = nums[i] + nums[left] + nums[right];if (sum == target) {return sum;}if (Math.abs(sum - target) < diff) {diff = Math.abs(sum - target);res = sum;}if (sum > target) {right--;} else {left++;}}}return res;    }}