LeetCode-172. Factorial Trailing Zeroes



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

题目描述

Given an integer

n

, return

the number of trailing zeroes in


n!

.

Note that

n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1

.

Example 1:

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

Example 2:

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

Example 3:

Input: n = 0
Output: 0

Constraints:

• 
0 <= n <= 10^4


解题思路

【C++】

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

【Java】

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