这个递归代码是如何工作的。我试图理解这个概念，但无法理解它

Question

import java.util.*;
public class HelloWorld {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int ar[] = new int[n];
        for (int i = 0; i < n; i++) {
            ar[i] = sc.nextInt();
        }
        int a = sc.nextInt();
        HelloWorld h = new HelloWorld();
        h.find(ar, a);
    }

    public void find(int ar[], int a) {
        System.out.println(a);
        for (int i = 0; i < ar.length; i++) {
            if (ar[i] == a) {
                //System.out.println(a+" "+i);
                a = a + 1;
                find(ar, a);
            }
        }
    }
}

Input:

2
1
2
1

Output:

1
2
3
3

Can anyone explains me how that recursive code generate the following output?

Answer 1

Don't get confused about the recursive code - AFAIK it could be done without recursive code.

It looks like it sorts the content of the variable ar in a weard kind and removes duplicates. But if the value does not exist in ar, it simply prints out a for each element of ar. And if a exists in ar, it prints the next higher value, and if that exists in ar, it prints the next higher value, but if currernt value of a does not exist in ar, just print it for the remaining loop iterations.

a at start = 0
result = 0 0 0 0

a at start = 1
result = 1 2 3 3

a at start = 2
result = 2 3 3 3

a at start = -infinite ... 0, 3, 4, ... +infite
result = a a a a

if ar = 1 2 4 5 (order of elements does not matter)
a = 0, result = 0 0 0 0
a = 1, result = 1 2 3 3
a = 2, result = 2 3 3 3
a = 3, result = 3 3 3 3
a = 4, result = 4 5 6 6
a = 5, result = 5 6 6 6
a = 6, result = 6 6 6 6

PS: The recusivity is not really needed, too few parameters change, only a.

The content of find without recursivity

public void find(int ar[], int a) {        
    for (int i = 0; i < ar.length; i++) {
        System.out.println(a);

        if (doesArrayContain(ar, a)) {
            //System.out.println(a+" "+i)
            a = a + 1;
        }
    }
}

private boolean doesArrayContain(int ar[], int a) {
    for (int idx = 0; idx < ar.length; idx++) {
        if (ar[idx] == a) {
            return true;
        }
    }
    return false;
}

Answer 2

it appears that you start with

looping an through an integer 'array of size n' to find a value 'a' if it exists you are then recusing to find 'a+1'

main.find [1,2], 1
console >>    1
    first recursive [1,2], 2.      --------  match for 1
    consoles >> 2
        second recursive [1,2] 3.  --------  match for 2
        consoles >> 3
        exit second recursive (no match for 3)
    third recursive [1,2], 2.      --------  match for 2
    consoles >> 3
    exit third recursive (no match for 3)
loop ends  

Answer 3

When you input

2
1
2
1

• n = the first number = 2
• ar = next n(2) numbers = [1,2]
• a = fourth number = 1

then the find method is called as:

h.find(ar = [1,2], a = 1);

For the output: Inside find method

• a = 1 is printed. As passed from the main method.
• ar = [1,2] is iterated.
• Since a[0] = 1 = a , a is incremented to 2. find(ar = [1,2], a = 2) is called.
  • a = 2 is printed.
  • ar = [1,2] is iterated.
  • This time a[0] = 1.= a = 2.
  • Next element, a[1] = 2 = a , a is incremented to 3. find(ar = [1,2], a = 3) is called.
    • a = 3 is printed. .
    • ar = [1,2] is iterated.
    • No element in ar is equal to a = 3 .
• a[1] = 2 = a . a was turned from 1 to 2 in prev iteration . Again a = 2+1=3 . find(ar = [1,2], a = 3) is called.
  • a = 3 is printed. .
  • ar = [1,2] is iterated.
  • No element in ar is equal to a = 3.

Final Output:

1
2
3
3

Point to note:

If you change value of primitive type variables (like int a ) inside a recursive call, the change will not affect value of a for the parent function.