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 / );
}
};