Trailing Zeros

 1 class Solution {
 2     /*
 3      * param n: As desciption
 4      * return: An integer, denote the number of trailing zeros in n!
 5      我们会发现： the number of 2s in prime factors is always more than or equal 
 6      to the number of 5s. So if we count 5s in prime factors, we are done. 
 7 
 8     How to count total number of 5s in prime factors of n!? A simple way is 
 9     to calculate floor(n/5). 
10 
11     问题转化为求阶乘过程中质因子5的个数，但是要注意25能提供2个5,125能提供3个5....
12     所以，count= floor(n/5) + floor(n/25) + floor(n/125) + ....
13      */
14 public long trailingZeros(long n) {
15         if (n < 0) return 0;
16 
17         long total = 0;
18         long base = 5;
19         while (base <= n) {
20             total += n / base;
21             base *= 5;
22         }
23         return total;
24     }
25 };