### 问题描述

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

### 代码

class Solution {
public:
int trailingZeroes(int n) {
int r = 0;
while (n) {
r += n / 5;
n /= 5;
}
return r;
}
};