Factorial Trailing Zeroes

Description

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 <= 104

Follow up: Could you write a solution that works in logarithmic time complexity?

Code

2 的因數永遠比 5 多，所以只要計算有幾個 5 就行了。

Time Complexity: $O(\log n)$, Space Complexity: $O(1)$

127 / 5 = 25

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

Source

• Factorial Trailing Zeroes - LeetCode