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LeetCode

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Problem

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

Note that $n! = n * (n - 1) * (n - 2) * … * 3 * 2 * 1$.

Example 1:

Input: n = 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:

Input: n = 5
Output: 1
Explanation: 5! = 120, one trailing zero.

Example 3:

Input: n = 0
Output: 0

Constraints:

  • $0 <= n <= 10^4$

Code

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