Factorial Trailing Zeroes

题目来源
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
计算n的阶乘的尾零数目。
我一开始想着是遍历一遍所有5和0的，然后依次相加，如下：

class Solution {
public:
    int trailingZeroes(int n) {
        int r = 0;
        for (int i=5; i<=n; i=i+5) {
            int cur = i;
            while (cur % 10 == 5 || cur % 10 == 0) {
                if (cur % 10 == 5) {
                    cur = cur / 5;
                    r++;
                }
                if (cur % 10 == 0) {
                    cur = cur / 10;
                    r++;
                }
            }
        }
        return r;
    }
};

但是这样做确实有点傻，TLE。得研究一下0尾零的增长规律。发现实现起来还是很简单的，要注意考虑溢出的情况。如下：

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