[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:

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

Java:

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);
    }
};

　　

 

　　

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