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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Have you met this question in a real interview?

思路 ：  cc150 上的原题， 查 5 的个数 ＋  25 的个数  ＋ 125 的个数

leetcode 用乘积会超时， 于是改成往下除。

public class Solution {    public int trailingZeroes(int n) {        if(n < 5)            return 0;        int ret = 0;        while(n > 0){            ret += n/5;            n /= 5;        }        return ret;    }}