LeetCode--No.172--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.

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思路：就是看有几个5的因子。 也是写了很多遍，逐渐填补上的漏洞。深深觉得自己bug free的能力欠缺的不是一星半点呢。

public class Solution { public int trailingZeroes(int n) { int m = n; int k = 0; int res = 0; while(m/5 > 0){ k++; m = m/5; } for(int i = 1; i <= k; i++){ res += n/(Math.pow(5,i)); } return res; } }