标签：
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
public class Solution { public int trailingZeroes(int n) { int sum=0; while(n/5!=0){ sum=sum+n/5; n=n/5; } return sum; } }
标签：
原文地址：http://www.cnblogs.com/mrpod2g/p/4248760.html
