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). 这道题让我们求最接近给定值的三数之和，是在之前那道的基础上又增加了些许难度，那么这道题让我们返回这个最接近于给定值的值，即我们要保证当前三数和跟给定值之间的差的绝对值最小，所以我们需要定义一个变量diff用来记录差的绝对值，然后我们还是要先将数组排个序，然后开始遍历数组，思路跟那道三数之和很相似，都是先确定一个数，然后用两个指针left和right来滑动寻找另外两个数，每确定两个数，我们求出此三数之和，然后算和给定值的差的绝对值存在newDiff中，然后和diff比较并更新diff和结果closest即可，代码如下：
class Solution {public: int threeSumClosest(vector & nums, int target) { int closest = nums[0] + nums[1] + nums[2]; int diff = abs(closest - target); sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size() - 2; ++i) { int left = i + 1, right = nums.size() - 1; while (left < right) { int sum = nums[i] + nums[left] + nums[right]; int newDiff = abs(sum - target); if (diff > newDiff) { diff = newDiff; closest = sum; } if (sum < target) ++left; else --right; } } return closest; }}; 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。