我无法弄清楚如何重置循环（参见示例）

Question

I need to write a method that accepts two ints as arguments, a min and a max. On the first line i need to print all numbers in that range (inclusive). On the next line I start with min+1, print all numbers up to max, and then go back to the front of the range and print min. Next line I start with min+2, and so on until I have repeated this starting with each number in the range.Very hard to explain, here's two examples: Say I pass 1 and 5 as the min and max arguments. I want the method to print this:

12345
23451
34512
45123
51234

Or if 3 and 9 were passed, I would expect this:

3456789
4567893
5678934
6789345
7893456
8934567
9345678

I've tried all kinds of things, I'm sure there is an easy way to do this that I am not realizing. I'm supposed to do this without arrays or arrayLists. I think I have a good base to work with, but I just can't figure out where to go from here. My base code prints this:

12345
2345
345
45
5

And this:

3456789
456789
56789
6789
789
89
9

I'm stumped. Here's my code:

public void printSquare(int min, int max){
   for (int i=min; i<=max; i++){
      for (int j=i; j<=max; j++){
         System.out.print(j);         
      }
   System.out.println();   
   }
}

Answer 1

This is a very simple implementation. Hope this helps!

   int n = max-min+1;
   for (int i=0 ; i<n; i++){
    for (int j=0; j<n; j++)
     cout<<min + (i+j)%n;
      cout<<"\n";
   }

Output:

min = 3 | max = 9 

    3456789
    4567893
    5678934
    6789345
    7893456
    8934567
    9345678

Answer 2

Here's the code..

 for i = 0 to max-min
 for j = 0 to max-min
 print min + (i+j)%n

Answer 3

You should think about how many values you want on each row, and then determine what those values should be. Its hard to make it any clearer without giving you the solution.

Let us know how you go.

Answer 4

Peter is right, and IMO is answering a homework question in the right manner. You know how many elements you want on each line, so you need an outer loop that gives you that many elements, this will stop you from getting the cascading behavior you're seeing now.

At that point you need to think about your inner loop(s), and you'll probably find this easiest using the modulus operator (%). This will allow you to iterate without ever going over your target.

You should be able to figure it out from there, and you're much better off figuring out the algorithm yourself than copying it from someone else, at least at this level of development. Good Luck!

Answer 5

public class Test1{
    public void printSquare(int min, int max){
        for (int i=min; i<=max; i++){
            for (int j=i; j<=max; j++){
                System.out.print(j);       
            }
            for(int k= min; k<i; k++){
                System.out.print(k);     
            }
            //System.out.print(i-1);
            System.out.println();   
        }
    }
    public static void main(String[] args){
        Test1 t = new Test1();
        t.printSquare(1,5);
    }
}

Answer 6

Think about a way to print the missing numbers. The answer is below, if you cannot come up with it you can check it.

---

This should also print the missing parts:

public void printSquare(int min, int max){  
   for (int i=min; i<=max; i++){  
      for (int j=i; j<=max; j++){  
         System.out.print(j);         
      }  
      for (int k=0; k<i-min; k++){  
         System.out.print(min+k);         
      }  
      System.out.println(); 
   }  
}

Answer 7

I didn't run this one but it might work:

public void printSquare(int min, int max){

   int dif = max - min;
   for (int i=min; i<=max; i++){

      for (int j=i; j <= i+dif ; j++){
         int temp = j;
         if ( temp > max ) temp = temp  - max;
         System.out.print(temp);         
      }

   System.out.println();   
   }

}

Answer 8

try and just shift an array like so:

   static void Main(string[] args)
    {
        // this will work equally well with numbers letters or other types of characters
        int[] nums = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        String a = "hello";
        for (int i = 0; i < nums.Length; i++)
        {
            int j = 5;
            int num = i;
            while (j-- > 0)
            {
                if (num >= nums.Length)
                {
                    num = 0;
                }

                // shift the loop
                Console.Write(nums[num++]);
            }
            Console.WriteLine();
        }
    }