The idea is that, if `n < 5`

then there's no zero, for `n = 5, 10, 15, 20`

there are 1,2,3,4 zeroes, which are `n / 5 + n / 5 / 5`

zeroes respectively. The same holds for `n = 25`

, there are `n / 5 + n / 5 / 5 = 5 + 1`

zeroes.

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