Problem Definition:
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
Solution： 这题有丰富的数学背景，代码很简单。。
1 def trailingZeroes( n): 2 c=0 3 while n>=5: 4 n/=5 5 c+=n 6 return c
一行写法：
1 def trailingZeroes( n): 2 return n/5+trailingZeroes(n/5) if n>=5 else 0
LeetCode#172 Factorial Trailing Zeroes
原文：http://www.cnblogs.com/acetseng/p/4658830.html