172. Factorial Trailing Zeroes

 

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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

这道题并没有什么难度,是让求一个数的阶乘末尾0的个数,也就是要找乘数中 10 的个数,而 10 可分解为2和5,而2的数量又远大于5的数量(比如1到 10 中有2个5,5个2),那么此题即便为找出5的个数。仍需注意的一点就是,像 25,125,这样的不只含有一个5的数字需要考虑进去,参加代码如下:

 

C++ 解法一:

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

 

Java 解法一:

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

 

这题还有递归的解法,思路和上面完全一样,写法更简洁了,一行搞定碉堡了。

 

C++ 解法二:

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

 

Java 解法二:

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

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/172

 

类似题目:

Number of Digit One

Preimage Size of Factorial Zeroes Function    

 

参考资料:

https://leetcode.com/problems/factorial-trailing-zeroes/

https://leetcode.com/problems/factorial-trailing-zeroes/discuss/52371/My-one-line-solutions-in-3-languages

https://leetcode.com/problems/factorial-trailing-zeroes/discuss/52373/Simple-CC%2B%2B-Solution-(with-detailed-explaination)

 

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