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

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

Example 1:

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

Example 2:

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

Note: Your solution should be in logarithmic time complexity.

给一个整数n，返回n的阶乘末尾0的个数。

找乘数中10的个数，而10可分解为2和5，而2的数量远大于5的数量，所以找出5的个数。

解法1：迭代Iterative

解法2: 递归Recursive

Java:

?

1 2 3 4 5 6 7 8 9 10

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

Java:

?

1 2 3 4 5

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

Python:

?

1 2 3 4 5 6 7 8

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

Python:

?

1 2 3 4 5 6 7

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++:

?

1 2 3 4 5 6 7 8 9 10 11

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

C++:

?

1 2 3 4 5 6

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

　　

 

　　

All LeetCode Questions List 题目汇总