[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.

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;
}
}

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

Number of Digit One

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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posted @ 2015-01-12 21:33  Grandyang  阅读(13037)  评论(4编辑  收藏  举报