Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
主要是思考清楚计算过程：
将一个数进行因式分解，含有几个5就可以得出几个0（与偶数相乘）。
代码很简单。
public class Solution {
public int trailingZeroes(int n) {
int result = 0;
long num = 5;
long div = (long) n;
while (div / num != 0){
result += div / num;
num *= 5;
}
return result;
}
}