LeetCode-in-Java
172. Factorial Trailing Zeroes
Medium
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 <= 104
Follow up: Could you write a solution that works in logarithmic time complexity?
Solution
public class Solution {
public int trailingZeroes(int n) {
int base = 5;
int count = 0;
while (n >= base) {
count += n / base;
base = base * 5;
}
return count;
}
}