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Factorial Trailing Zeroes

2024-10-19 01:02:01   209人阅读

 

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 num15      * @return16      */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 num33      * @return34      */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 }

 

Factorial Trailing Zeroes


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