问题：
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). 题意：给定一个有n个整数的数组S，找出S中3个数，使其和等于一个给定的数，target。
返回这3个数的和，你可以假定每个输入都有且只有一个结果。
例如，给定S = {-1 2 1 -4}，和target = 1。那么最接近target的和是2。(-1 + 2 + 1 = 2)。
在这个题目中，我们要做以下几件事：
用 sort 对输入的数组进行排序
求出长度 len ， current 之所以要小于 len−2 ，是因为后面需要留两个位置给 front 和 back
始终保证 front 小于 back
计算索引为 current 、 front 、 back 的数的和，分别有比 target 更小、更大、相等三种情况
更小：如果距离小于 close ，那么 close 便等于 target−sum ，而结果就是 sum 。更大的情况同理
如果相等，那么要记得将 0 赋值给 close ， result 就直接等于 target 了
随后为了避免计算重复的数字，用三个 do/while 循环递增或递减它们
代码： class Solution { public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); int len = nums.size(); int result = INT_MAX, close = INT_MAX; for (int current = 0; current < len - 2; current++) { int front = current + 1, back = len - 1; while (front < back) { int sum = nums[current] + nums[front] + nums[back]; if (sum < target) { if (target - sum < close) { close = target - sum; result = sum; } front++; } else if (sum > target) { if (sum - target < close) { close = sum - target; result = sum; } back--; } else { close = 0; result = target; do { front++; } while (front < back&&nums[front - 1] == nums[front]); do { back--; } while (front < back&&nums[back + 1] == nums[back]); } } while (current < len - 2 && nums[current + 1] == nums[current]) { current++; } } return result; } };