[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 = 0;
        while (n > 0) {
            res += n / 5;
            n /= 5;
        }
        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 == 0 ? 0 : n / 5 + trailingZeros(n / 5);
    }
};