魔术火柴问题极客为极客面试问题

Question

我在极客中接受了一次对极客的采访，在那里我遇到了一个魔术火柴棒问题，我没能及时解决。一天后，我自己解决了这个问题，在堆栈溢出的情况下，我也没能找到解决方案，所以如果您遇到这个问题，这里有一个解决方案：

//magic match-stick problem
/*in this problem set we are given n number of match boxes, 
* you need to add matches in a match that is 
* the n number of matches in each box,
* but the catch is instead of adding the matches, 
* you need to subtract the matches pair that you will be inserting in place 
* of every new match stick until there is only last box left with final match stick score
* eg: 1st match box: 4match sticks: 1,2,3,4 match numbers
* 2nd time: 2 match sticks : (1-2), (3-4)=> 1, 1
* last time: (1-1) => 0
* 
* I know I cannot explain the problem correctly in English, but I will be able to explain it to you in hindi v well
* also if you can understand the code, you will understand the question
* this took me one day to solve, many minor mistakes I did
* this is a geeks for geeks interview question
* question name magical matchsticks*/

public class practice{
    public static void main(String args[]) {    
        int a[] = {1,2,3,4,5,6};
        int num;
        int count;
        int n = a.length;
        
        do
        {
    
            num=0;
            count=0;
            
            if(n==1)
            {
                System.out.println(a[0]+"last");
            }
            
            else if(n>1)
            {
                
                
                if(n%2==0)
                {
                    n=n/2;
                    
                    for(int i = 0; i<((n)) ; i++)
                    {
                        
                        a[i] = a[num]-a[num+1];
                        
                        if(a[i]<0)
                        {
                            a[i] = a[i]*(-1);
                            System.out.print(a[i]+" "+a[i + 1]);
                        }
                        
                        System.out.println(" a ");
                        
                        num = num + 2;
                        count++;
                        
                    }
                }
                else if(n%2==1)
               {
                    n = (n/2) +1;
                    for(int i = 0; i<(n); i++)
                   {    
                        a[i] = a[num]-a[num+1];
                        if(a[i]<0)
                        {
                            a[i] = a[i]*(-1);
                            System.out.print(a[i]+" "+a[i + 1]);
                        }
                        System.out.println("b ");
                        num = num + 2;
                        count++;
                    }
                    
                }
            }
            System.out.println(n);
            for(int j = 0; j<a.length;j++) 
            {
                System.out.print(a[j]);
            }
            System.out.println(" ");
                
        }while(n>1); 
        
        
        for(int k = 0; k<a.length;k++) 
        {
            System.out.print(a[k]);
        }
        System.out.println(" ");
        
        
    }
}

这个问题将显示程序进行过程中的所有步骤，您将需要进行更改，以便从一个单独的类或函数返回值。

我将开始为这个帐户上的竞争性编程提供解决方案。

Answer 1

这个问题是用上面的答案来解决的。

在我的github页面上也提供了答案：