# [LeetCode] 172. Factorial Trailing Zeroes 求阶乘末尾零的个数

2021年09月15日 阅读数：1

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

Example 1:java

```Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.```

Example 2:python

```Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.```

Note: Your solution should be in logarithmic time complexity.post

Java:get

```public class Solution {
public int trailingZeroes(int n) {
int res = 0;
while (n > 0) {
res += n / 5;
n /= 5;
}
return res;
}
}　　```

Java:it

```public class Solution {
public int trailingZeroes(int n) {
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
}
}
```

Python:

```class Solution:
# @return an integer
def trailingZeroes(self, n):
result = 0
while n > 0:
result += n / 5
n /= 5
return result　　```

Python:

```class Solution(object):
def trailingZeroes(self, n):
"""
:type n: int
:rtype: int
"""
return 0 if n == 0 else n / 5 + self.trailingZeroes(n / 5)   　　```

C++:

```class Solution {
public:
int trailingZeroes(int n) {
int res = 0;
while (n) {
res += n / 5;
n /= 5;
}
return res;
}
};
```

C++:

```class Solution {
public:
int trailingZeroes(int n) {
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
}
};
```