LeetCode(31)-Factorial Trailing Zeroes

题目：

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

思路：

代码：

class Solution {
/*
* param n: As desciption
* return: An integer, denote the number of trailing zeros in n!
*/
public long trailingZeros(long n) {
// write your code here
return n / 5 == 0 ? 0 : n /5 + trailingZeros(n / 5);
}
};

暴力方法：（不推荐）

public class Solution {
public int trailingZeroes(int n) {
int count = 0;
if(get(n) == 0){
return 1;
}
int sum = get(n);
while(sum > 0){
if(sum % 10 == 0){
count++;
}
sum = sum /10;
}
return sum;
}
public int get(int n){
int all = 0;
if(n == 0){
return 0;
}
for(int i = 1;i <= n;i++){
all = all*i;
}
return all;
}
}