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).
问题：给定一个数组和一个整数 n ，求数组中的三个元素，他们的和离 n 最近。
这道题和 3Sum 很相似，解题思路也很相似，先排序，然后采用双指针法依次比较。
由于 3Sum 很相似，思路就不详说。主要说下不同点：
- 由于是求最接近值，当 s[i] + s[l] + s[r] - target == 0 的时候，即可返回结果。
- 返回的结果是最接近 n 的三个元素之和。
int threeSumClosest(vector<int>& nums, int target) {
std::sort(nums.begin(), nums.end());
int closest = target - nums[] - nums[] - nums[];
for (int i = ; i < nums.size(); i++) {
int l = i + ;
int r = (int)nums.size() - ;
int newTarget = target - nums[i];
while (l < r) {
if (nums[l] + nums[r] == newTarget) {
return target;
}
int diff = newTarget - nums[l] - nums[r];
if (abs(diff) < abs(closest)) {
closest = diff;
}
if (nums[l] + nums[r] < newTarget){
l++;
}else{
r--;
}
}
}
return target - closest;
}