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)

　　

 package com.lun.alrithmetic;
 /*
  * Q1: what's the primary factor? (1 2 3)
  * Q2: i & i+2ne能否遍历出所有质数
  */

 public class LargestPrimeFactor {

     public static void main(String[] args) {
         // TODO Auto-generated method stub
         long n = 600851475143L;
         System.out.println("The largest prime factor of the number '600851475143' is "+largestPrimeFactor(n));
     }

     public static boolean isPrime(long n){
         if(n<4)
             return n>1;
         if(n%2==0 || n%3==0)
             return false;
         for(long i=5; i*i<n; i+=6)
             if(n%i==0 || n%(i+2)==0)
                 return false;
         return true;
     }

     public static long largestPrimeFactor(long n){
             if(n%2==0){
                 while(n%2==0)
                     n /= 2;
             }
             if(n%3==0){
                 while(n%3==0)
                     n /= 3;
             }
             for(int i=5; i*i<n; i+=6){
                 if(n%i==0){
                     while(n%i==0)
                         n /= i;
                 }
                 int j = i+2;
                 if(n%j==0){
                     while(n%j==0)
                         n /= j;
                 }
             }
         return n;
     }
 }

way1

 package com.lun.alrithmetic;
 /*
  * Question: What's the largest prime factor of the number '600851475143' ?
  * Function: Divide & Conquer, analyse prime factor
  */

 public class LargestPrimeFactor2 {

     public static void main(String[] args) {
         // TODO Auto-generated method stub
         long n = 600851475143L;
         System.out.println("The largest prime factor of the number '600851475143' is "+largest(n));
     }

     public static boolean isPrime(long n){
         if(n<4)
             return n>1;
         if(n%2==0 || n%3==0)
             return false;
         for(long i=5; i*i<n; i+=6)
             if(n%i==0 || n%(i+2)==0)
                 return false;
         return true;
     }

     public static long largest(long n){
         if(isPrime(n))
             return n;
         else{
             if(n%2==0)
                 return largest(n/2);
             if(n%3==0)
                 return largest(n/3);
             for(long i=5; i*i<n; i+=6){
                 if(n%i==0)
                     return largest(n/i);
                 if(n%(i+2)==0)
                     return largest(n/(i+2));
             }
         }

         return 1;
     }

 }

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。