C++实现LeetCode(172.求阶乘末尾零的个数)

时间:2022-09-15 11:18:53

[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.

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:

?
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;
    }
};

Java 解法一:

?
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;
    }
}

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:

?
1
2
3
4
5
6
class Solution {
public:
    int trailingZeroes(int n) {
        return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
    }
};

Java 解法二:

?
1
2
3
4
5
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)

到此这篇关于C++实现LeetCode(172.求阶乘末尾零的个数)的文章就介绍到这了,更多相关C++实现求阶乘末尾零的个数内容请搜索服务器之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持服务器之家!

原文链接:https://www.cnblogs.com/grandyang/p/4219878.html