Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

public class Solution {
    public int trailingZeroes(int n) {
        int total = 0;
        int base = 5;
        while(n>=base) {
            total += n/base;
            if(base>Integer.MAX_VALUE/5) break;
            base *= 5;
        }
        return total;
    }
}

5倍数的个数，25倍数的个数，125倍数的个数，725倍数的个数。。。

=====

public class Solution {
    public int trailingZeroes(int n) {
        int res = 0;
        int fives = 5;
        int max = Integer.MAX_VALUE / 5;
        while (true) {
            int num = n / fives;
            if (num == 0) break;
            res += num;
         
            if (max < fives) break;
            fives *= 5;
        }
        return res;
    }
}

=======

public class Solution {
    public int trailingZeroes(int n) {
        int res = 0;
        if (n <= 0) {
            return res;
        }
        while (n != 0) {
            res += n / 5;
            n /= 5;
        }
        return res;
    }
}