Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
一刷
题解：
求n!尾部有多少个0。其实主要就是看有多少个5，有多少个5就有多少个0
count = n / 5 + n / 25 + n / 125 + ....
如果有25，会产生2个0， 125会产生3个0
public class Solution {
public int trailingZeroes(int n) {
if(n<=0) return 0;
int res = 0;
while(n>0){
res += n/5;
n/=5;
}
return res;
}
}
