Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
此题是求阶乘后面零的个数。
public class Solution {
public int trailingZeroes(int n) {
int t=0;
while(n!=0){
n/=5;
t+=n;
}
return t;
}
}