LeetCode 172 Factorial Trailing Zeroes

Posted 2020-08-06

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

Note: Your solution should be in logarithmic time complexity.

 

思路：

首先末尾0的个数取决于素因数中2和5个数较少者，其次5的个数要远远小于2的个数，因此返回素因数中5的个数即可。有公式[n/5] + [n/25] + [n/125] + ... + [n/5^x]

 

解法：

 1 public class Solution
 2 {
 3     public int trailingZeroes(int n)
 4     {
 5         int numberOfFive = 0;
 6 
 7         while(n / 5 > 0)
 8         {
 9             n /= 5;
10             numberOfFive += n;
11         }
12 
13         return numberOfFive;
14     }
15 }