[LintCode]Trailing Zeroes末尾ゼロの個数

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の原題は、私の前のブログFactorial Trailing Zeroesを参照してください.
 
解法1:

class Solution {
 public:
    // param n : description of n
    // return: description of return 
    long long trailingZeros(long long n) {
        long long res = 0;
        while (n > 0) {
            res += n / 5;
            n /= 5;
        }
        return res;
    }
};

 
解法2:

class Solution {
 public:
    // param n : description of n
    // return: description of return 
    long long trailingZeros(long long n) {
        return n == 0 ? 0 : n / 5 + trailingZeros(n / 5);
    }
};