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
1 public class Solution {
2 public int trailingZeroes(int n) {
3 int numsOf2 = 0;
4 int numsOf5 = 0;
5 for(int i = 2; i <= n; i++){
6 numsOf2 += get2nums(i);
7 numsOf5 += get5nums(i);
8 }
9 return numsOf2 < numsOf5 ? numsOf2 : numsOf5;
10 }
11
12 /**
13 * 计算num中2作为乘积因子的个数
14 * @param num
15 * @return
16 */
17 private int get2nums(int num){
18 int numsOf2 = 0;
19 if(num % 2 != 0)
20 return 0;
21 else{
22 while(num % 2 == 0){
23 num = num / 2;
24 numsOf2 ++;
25 }
26 }
27 return numsOf2;
28 }
29
30 /**
31 * 获取5的个数
32 * @param num
33 * @return
34 */
35 private int get5nums(int num){
36 int numsOf5 = 0;
37 if(num % 5 != 0)
38 return 0;
39 else{
40 while(num % 5 == 0){
41 num = num / 5;
42 numsOf5 ++;
43 }
44 }
45 return numsOf5;
46 }
47 }
