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:
1 <= n <= 104
public class Solution {
public int TrailingZeroes(int n) {
if(n==0){
return 0;
}
return n/5 + TrailingZeroes(n/5);
}
}Time Complexity: O(n)
Space Complexity: O(1)
Because all trailing 0 is from factors 5 * 2.
But sometimes one number may have several 5 factors, for example, 25 have two 5 factors, 125 have three 5 factors. In the n! operation, factors 2 is always ample. So we just count how many 5 factors in all numbers from 1 to n.
