LeetCode | 3Sum Closest（找到三个数使其和与target最接近）

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).

题目解析：

找三个数，使其和与target最近。

一开始也是按照先选一个a，然后让target-a为要找的数据，进入子函数去循环。还想着什么时候要退出，是否要遍历完i+1...n个数据。

由于数的变化性比较大，我们找到i和j的差值与target-a相近了，但i和j之间的值可能更近，也可能更远。因此要遍历完全，并且没遍历一步都要与target-a比较。

我们干脆直接维持一个和sum = a+b+c。让sum-target变小时，就更新minSum。而左右指针也是同样的走向，当sum-target>0,end--；反之，begin++。

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
if(num.size() < 3)
return -1;
sort(num.begin(),num.end());
int min = num[0]+num[1]+num[2]-target;
for(int i = 0;i < num.size()-1;i++){
int begin = i+1;
int end = num.size()-1;
while(begin < end){
int sum = num[i]+num[begin]+num[end]-target;
if(sum==0) return target;
if(abs(sum) < abs(min)) min = sum;
if(sum<0)
begin++;
else
end--;
}
}
return min+target;
}
};