Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
此题是求阶乘后面零的个数。
public class Solution { public int trailingZeroes(int n) { int t=0; while(n!=0){ n/=5; t+=n; } return t; } }
每天一道LeetCode--172. Factorial Trailing Zeroes
原文：http://www.cnblogs.com/xiaoduc-org/p/6082097.html