Given an integer n, return the number of trailing zeroes in n!.
Follow up: Could you write a solution that works in logarithmic time complexity?
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 <= 104struct Solution;
impl Solution {
fn trailing_zeroes(mut n: i32) -> i32 {
let mut sum = 0;
while n > 0 {
sum += n / 5;
n /= 5;
}
sum
}
}
#[test]
fn test() {
assert_eq!(Solution::trailing_zeroes(3), 0);
assert_eq!(Solution::trailing_zeroes(5), 1);
assert_eq!(Solution::trailing_zeroes(10), 2);
}