Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

只有2 * 5才会产生0只要计算出因子中2和5的个数取小的的那个就好了

 public class Solution {
     public int trailingZeroes(int n) {
         int numsOf2 = 0;
         int numsOf5 = 0;
         for(int i = 2; i <= n; i++){
             numsOf2 += get2nums(i);
             numsOf5 += get5nums(i);
         }
         return numsOf2 < numsOf5 ? numsOf2 : numsOf5;
     }

     /**
      * 计算num中2作为乘积因子的个数
      * @param num
      * @return
      */
     private int get2nums(int num){
         int numsOf2 = 0;
         if(num % 2 != 0)
             return 0;
         else{
             while(num % 2 == 0){
                 num = num / 2;
                 numsOf2 ++;
             }
         }
         return numsOf2;
     }

     /**
      * 获取5的个数
      * @param num
      * @return
      */
     private int get5nums(int num){
         int numsOf5 = 0;
         if(num % 5 != 0)
             return 0;
         else{
             while(num % 5 == 0){
                 num = num / 5;
                 numsOf5 ++;
             }
         }
         return numsOf5;
     }
 }