LeetCode //C - 172. Factorial Trailing Zeroes

172. Factorial Trailing Zeroes

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:

• $0<=n<=10^4$

From: LeetCode
Link: 172. Factorial Trailing Zeroes

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Solution:

Ideas：

• We initialize count to 0 to keep track of the number of 5s.
• We use a while loop that runs as long as n is greater than 0.
• Inside the loop, we divide n by 5 to find out how many multiples of 5, 25, 125, etc., are there in n.
• We add this value to count.
• Finally, we return count, which represents the number of trailing zeroes in n!.

Code:

int trailingZeroes(int n) {
    int count = 0;
    while (n > 0) {
        n /= 5;
        count += n;
    }
    return count;
}
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