Question

Formatted question description: https://leetcode.ca/all/172.html

 172	Factorial Trailing Zeroes

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

 Example 1:

 Input: 3
 Output: 0
 Explanation: 3! = 6, no trailing zero.
 Example 2:

 Input: 5
 Output: 1
 Explanation: 5! = 120, one trailing zero.
 Note: Your solution should be in logarithmic time complexity.

 @tag-math

Algorithm

To find the number of 10 in the multiplier, and 10 can be decomposed into 2 and 5, and the number of 2 is much greater than the number of 5 (for example, there are 2 5s in 1 to 10, 5 2s), then this question is find the counts of 5.

One thing that still needs to be noted is that numbers like 25, 125, which contain more than a 5, need to be taken into consideration.

Code

public class Factorial_Trailing_Zeroes {

    public static void main(String[] args) {
        Factorial_Trailing_Zeroes out = new Factorial_Trailing_Zeroes();
        Solution_iteration s = out.new Solution_iteration();

        s.trailingZeroes(126);
//        System.out.println(s.trailingZeroes(126));
    }


    public class Solution_iteration {
        public int trailingZeroes(int n) {
            int res = 0;
            while (n > 0) {
                res += n / 5;
                n /= 5;

                /*
                System.out.println(res);
                    25
                    30
                    31
                    31
                */
            }
            return res;
        }
    }

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

############

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

class Solution:
    def trailingZeroes(self, n: int) -> int:
        ans = 0
        while n:
            n //= 5
            ans += n
        return ans

############

class Solution(object):
  def trailingZeroes(self, n):
    """
    :type n: int
    :rtype: int
    """
    count, k = 0, 5
    while n:
      k = n / 5
      count += k
      n = k
    return count

﻿class Solution {
public:
    int trailingZeroes(int n) {
        int ans = 0;
        for (int i = 5; i <= n; i *= 5) ans += n / i;
        return ans;
    }
};

func trailingZeroes(n int) int {
	ans := 0
	for n > 0 {
		n /= 5
		ans += n
	}
	return ans
}

function trailingZeroes(n: number): number {
    let ans = 0;
    while (n > 0) {
        n = Math.floor(n / 5);
        ans += n;
    }
    return ans;
}

172 - Factorial Trailing Zeroes

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