Factorial Trailing Zeroes

Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

------------------------------ thinking -----------------------------------------------

the number of 0 is decided by how many 5 it has in the array, because the number of 2 is always bigger than 5.

----------------------------- mistakes -----------------------------------------------

When I firstly implemented it, I think of 5 and 10 separately, which is obviously wrong. Because 10 is constructed by 2 & 5!

---------------------------- codes ---------------------------

class Solution {

public:

    int trailingZeroes(int n) {

        int result = 0;

        int cur_num = n;

        while (cur_num > 0) {

            result += cur_num / 5;

            cur_num /= 5;

        }

        return result;

    }

};