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.

https://leetcode.com/problems/factorial-trailing-zeroes/

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package leetcode;
/*因为10=2*5,所以0的个数就是100!因式分解后2*5(必须配对)的个数.显然因子中2的个数比5多,因此问题划归为求解100!的因子中5的个数.
而1 2 3 4.....100这一百个数中5 10 15 20 .....100各含一个因子5(总共20个),25 50 75 100各含两个
(因为前面已经统计过一次,只能算一个)
这样答案就是20+4=24个0.*/
public class Solution 
{
    public static int trailingZeroes(int n) 
    {
        int cout=0;
        if(n==0)
            return cout;
        for(int i=1;n>=Math.pow(5,i);i++)
        {
            cout+=n/Math.pow(5,i);
        }
        return cout;
    }
    public static void main(String args[])
    {
        System.out.println(trailingZeroes(30));
    }
}