Factorial Trailing Zeroes
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.
对n!做质因数分解n!=2x*3y*5z*...
显然0的个数等于min(x,z)，并且min(x,z)==z
解法一：
从1到n中提取所有的5
class Solution { public: int trailingZeroes(int n) { int ret = 0; for(int i = 1; i <= n; i ++) { int tmp = i; while(tmp%5 == 0) { ret ++; tmp /= 5; } } return ret; } };
解法二：
由上述分析可以看出，起作用的只有被5整除的那些数。能不能只对这些数进行计数呢？
存在这样的规律：[n/k]代表1~n中能被k整除的个数。
因此解法一可以转化为解法二
class Solution { public: int trailingZeroes(int n) { int ret = 0; while(n) { ret += n/5; n /= 5; } return ret; } };
【LeetCode】Factorial Trailing Zeroes (2 solutions)
原文：http://www.cnblogs.com/ganganloveu/p/4193373.html