The largest prime factor（最大质因数）

1. 问题:

The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?

2. 解法(by java in Eclipse)

 1 package com.lun.alrithmetic;
 2 /*
 3  * Q1: what‘s the primary factor? (1 2 3)
 4  * Q2: i & i+2ne能否遍历出所有质数
 5  */
 6
 7 public class LargestPrimeFactor {
 8
 9     public static void main(String[] args) {
10         // TODO Auto-generated method stub
11         long n = 600851475143L;
12         System.out.println("The largest prime factor of the number ‘600851475143‘ is "+largestPrimeFactor(n));
13     }
14
15     public static boolean isPrime(long n){
16         if(n<4)
17             return n>1;
18         if(n%2==0 || n%3==0)
19             return false;
20         for(long i=5; i*i<n; i+=6)
21             if(n%i==0 || n%(i+2)==0)
22                 return false;
23         return true;
24     }
25
26     public static long largestPrimeFactor(long n){
27             if(n%2==0){
28                 while(n%2==0)
29                     n /= 2;
30             }
31             if(n%3==0){
32                 while(n%3==0)
33                     n /= 3;
34             }
35             for(int i=5; i*i<n; i+=6){
36                 if(n%i==0){
37                     while(n%i==0)
38                         n /= i;
39                 }
40                 int j = i+2;
41                 if(n%j==0){
42                     while(n%j==0)
43                         n /= j;
44                 }
45             }
46         return n;
47     }
48 }

way1

 1 package com.lun.alrithmetic;
 2 /*
 3  * Question: What‘s the largest prime factor of the number ‘600851475143‘ ?
 4  * Function: Divide & Conquer, analyse prime factor
 5  */
 6
 7 public class LargestPrimeFactor2 {
 8
 9     public static void main(String[] args) {
10         // TODO Auto-generated method stub
11         long n = 600851475143L;
12         System.out.println("The largest prime factor of the number ‘600851475143‘ is "+largest(n));
13     }
14
15     public static boolean isPrime(long n){
16         if(n<4)
17             return n>1;
18         if(n%2==0 || n%3==0)
19             return false;
20         for(long i=5; i*i<n; i+=6)
21             if(n%i==0 || n%(i+2)==0)
22                 return false;
23         return true;
24     }
25
26     public static long largest(long n){
27         if(isPrime(n))
28             return n;
29         else{
30             if(n%2==0)
31                 return largest(n/2);
32             if(n%3==0)
33                 return largest(n/3);
34             for(long i=5; i*i<n; i+=6){
35                 if(n%i==0)
36                     return largest(n/i);
37                 if(n%(i+2)==0)
38                     return largest(n/(i+2));
39             }
40         }
41
42         return 1;
43     }
44
45 }

way2

注：利用分解质因数的方法，从小向大用质数整除（如果此质数恰好是n的因数的话）n,即不断的减小n的规模，最后即可求的最大质因数。

法二，采用了递归的方法易于理解，但是内存消耗较大；建议用法一。