Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

 

Solution1:

class Solution {
public:
    int trailingZeroes(int n) {
        long int temp = factorial(n);
        int count = 0;
        bool finish = false;
        while(!finish){
            if(temp % 10 == 0){
                count++;
                temp /= 10;
            }
            else finish = true;
        }
        return count;
    }
    long int factorial(int n){
        if(n == 0) return 1;
        else return n*factorial(n-1);
    }
};

结果显示超时，所以只能换个思维。每当出现5的时候，由于在这之前已经出现了偶数，所以可以产生1个0。遇到25、50、75、100、125等25的倍数时，0的个数相应增加。当在草稿纸上每5个数写出5~125的output时，可以发现很明显的规律，如以下代码：

Solution2:

class Solution {
public:
    int trailingZeroes(int n) {
        if(n == 0) return 0;
        
        int count = 0;
        while(n >= 5){
            count += n / 5;
            n /= 5;
        }
        return count;
    }
};