Question

172. Factorial Trailing Zeroes

中文文档

Description

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?

Solutions

Solution 1

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

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

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