[LintCode] Trailing Zeroes 末尾零的个数

Write an algorithm which computes the number of trailing zeros in n factorial.

Have you met this question in a real interview?

Yes

Example

11! = 39916800, so the out should be 2

Challenge

O(log N) time

LeetCode上的原题，请参见我之前的博客Factorial Trailing Zeroes。

解法一：

class Solution {
 public:
    // param n : description of n
    // return: description of return
    long long trailingZeros(long long n) {
        long long res = ;
        while (n > ) {
            res += n / ;
            n /= ;
        }
        return res;
    }
};

解法二：

class Solution {
 public:
    // param n : description of n
    // return: description of return
    long long trailingZeros(long long n) {
        return n ==  ?  : n /  + trailingZeros(n / );
    }
};