LeetCode - Factorial Trailing Zeroes

Factorial Trailing Zeroes

2015.1.23 18:46

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

Solution:

　　The number of zeros is simply the number of 5s. That's it.

　　Time complexity is O(log(n)), space complexity is O(1).

Accepted code:

 1 // 1AC, log5(N)

 2 class Solution {

 3 public:

 4     int trailingZeroes(int n) {

 5         int sum;

 6         

 7         sum = 0;

 8         while (n > 0) {

 9             sum += n / 5;

10             n /= 5;

11         }

12         

13         return sum;

14     }

15 };