LeetCode - Factorial Trailing Zeroes

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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 };

LeetCode - Factorial Trailing Zeroes

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原文地址：http://www.cnblogs.com/zhuli19901106/p/4244886.html