172. Factorial Trailing Zeroes

n!就是1*2*3*4*.......*n，没出现一个5，0就增加一个，比如，103！=1*2*3*4*5*6*7*........*10*.........*103,所所以至少出现5，10，15，20.。。。。100，除以5又得，1，2，3，4，5，。。。。20这又是四个，依次往下，即可

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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public class Solution {
    public int trailingZeroes(int n) {
        int result = 0;
        if(n <= 4){
            return 0;
        }
        while(n/5 != 0){
            n/=5;
            result += n;
        }
        return result;
    }
}