Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
思路：编程之美里有，就是找因子5的个数。
int trailingZeroes(int n) { int ans = 0; while(n > 0) { ans += n / 5; n /= 5; } return ans; }
【leetcode】Factorial Trailing Zeroes（easy）
原文：http://www.cnblogs.com/dplearning/p/4434846.html