172. Factorial Trailing Zeroes

https://leetcode.com/problems/factorial-trailing-zeroes/description/
Solution. The number of trailing 0 of n! equals the total number of factor 5 in 1, 2, ... , n. 
Each step, n/5 is the number of integers that have at least one factor of 5 in 1,2,...,n.
    int trailingZeroes(int n) {
        if(n == 0) return 0;
        int nz = 0;
        while(n) {
            nz += n/5;
            n /= 5;
        }
        return nz;
    }