Generate 10 numbers and move first number to the end 10 times

Question

I have code that generates 1 to 10 numbers:

for (int i = 0; i <=10; i++)
{
     for (int j = 1; j <=10; j++)
     {
         int respones = i;
         int respones1 = j;

         if (respones1 > respones)
         {
              text.append(String.valueOf(respones1));
         }
     }
}

I get this result:

12345678910
2345678910
345678910
45678910
5678910
678910
78910
8910
910
10

But, I want this result:

12345678910
23456789101
34567891012
45678910123
56789101234
67891012345
78910123456
89101234567
91012345678
10123456789

How can I get my code so that it moves the first number to the end of the string?

Answer 1

Try this one:

for (int i = 0; i < 10; i++) {
    for (int j = i + 1; j <= i + 10; j++) {
        int respones = i;
        int respones1 = j;
        if (respones1 > respones) {
            text.append(String.valueOf(respones1 > 10 ? respones1 % 10 : respones1));
        }
    }
}

Answer 2

I believe the best way to approach this would be to use a deque/queue as it makes rotating the numbers in the sequence achieveable in just two operations.

ArrayDeque<int> deque = new ArrayDeque<int>(10); // Create a queue 
for(int i = 1; i <= 10; ++i) // Fill queue with [1..10]
{
    deque.addLast(i);
}

StringBuilder builder = new StringBuilder(); // A StringBuilder will be more efficient than string concatenation
for(int i = 1; i <= 10; ++i)
{
    for(int item : deque) // Append the queue contents to the string
    {
        builder.append(item);
    }
    builder.append("\n"); // New line

    int temp = deque.removeFirst(); // Take the first item from the deque
    deque.addLast(temp); // And append it to the end of the deque
}