題目

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

Note: Your solution should be in logarithmic time complexity.

Credits: Special thanks to @ts for adding this PRoblem and creating all test cases.

Subscribe to see which companies asked this question.

思路

計算n的階乘的最后有幾個零，就是計算因數(shù)中2和5的最小個數(shù)，5的個數(shù)肯定比2的少，就是計算因數(shù)中5的個數(shù)

代碼