在这种情况下，递归到底如何工作？

Question

I am new to this forum and I hope that I don't make any mistakes. Here's my question. I think I understand the basic principle of recursion and its implementation, yet I have problems to understand the EXACT steps the computer is doing when executing a recursive code.

Here's an example:

public class Main {

public static void MyAlgorithm (int [] A, int l, int r ) {
    if (l<r) {
        int m = (l+r)/2;
        MyAlgorithm(A, l, m);

        for (int i = l; i<=r; i++) {
            System.out.println(A[i]);
        }

        MyAlgorithm(A, m+1, r);
    }


}
public static void main (String args[]) {

    int [] A = {1,2,3,4};

    MyAlgorithm (A, 0, 3);
}

}

The output is: 12123434

But how does this exactly work? At first, the code calculates m as 2. It then performs the method again on the subarray (1,2) and puts this on a stack. Then it does it again, takes the half and puts 1 on the stack. As far as I understand the issue, it then stops, pops 1 from the stack and prints it. But when does it put the 2 on the stack separately? And when exactly does it start the recursion part with the other half? I simply cannot really understand how the computer is executing the single steps and how it "knows" when to start calling the other half of the array recursively. I hope you understand my question and I am grateful for every response!

Thanks in advance :)

Answer 1

Execution call will be something like this

MyAlgorithm(A,0,3) 
    (l < r : true)
     m = (0+3)/2=1
     MyAlgorithm(A,0,1) 
            (l < r : true) 
            m = (0+1)/2=0
            MyAlgorithm(A,0,0) 
                (l < r : false) return

            loop from l=0 to r=1
            System.out.println(A[0]);
            System.out.println(A[1]);

            //MyAlgorithm(A,m+1,r)
            MyAlgorithm(A,1,1) 
                (l < r : false) return

    loop from 1=0 to r=3
    System.out.println(A[0]);
    System.out.println(A[1]);
    System.out.println(A[3]);
    System.out.println(A[4]);

    //MyAlgorithm(A,m+1,r)
    MyAlgorithm(A,2,3)
        (l < r : true)
        m=(2+3)/2 = 2
        MyAlgorithm(A,2,2)
            (l < r : false) return

    loop from 1=2 to r=3
    System.out.println(A[2]);
    System.out.println(A[3]);

    MyAlgorithm(A,3,3)
            (l < r : false) return      
return

This leads to answer: 12123434