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).分析：
先排序，然后左右夹逼，复杂度O(n^2)。
代码：
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {int result = 0;int min_gap = INT_MAX;sort(nums.begin(), nums.end());for (auto a = nums.begin(); a != prev(nums.end(), 2); ++a) {auto b = next(a);auto c = prev(nums.end());while (b < c) {const int sum = *a + *b + *c;const int gap = abs(sum - target);if (gap < min_gap) {result = sum;min_gap = gap;}if (sum < target) ++b;else --c;}}return result;}
};