递归中的ArrayLists的ArrayList

Question

I've been working on a larger project, and ran into a problem which I have replicated here in a much simpler fashion. What I'm trying to do is to add an ArrayList of Integers into another ArrayList. The problem is that every ArrayList I add into the larger ArrayList gets updated as if they were all the same.

public class RecursionTest {
static ArrayList<Integer> test = new ArrayList<Integer>();
static ArrayList<ArrayList<Integer>> test1 = new ArrayList<ArrayList<Integer>>();

public static void testRecurse(int n) {
    test.add(n);
    if (n % 2 == 0) {
        test1.add(test);
    }
    if (n == 0) {   
        for (ArrayList<Integer> a : test1) {
            for (Integer i : a) {
                System.out.print(i + "  ");
            }
            System.out.println();
        }
        return;
    }       
    testRecurse(n - 1);     
}

public static void main(String[] args) {        
    testRecurse(10);
}
}

The output I get is:

10  9  8  7  6  5  4  3  2  1  0  
10  9  8  7  6  5  4  3  2  1  0  
10  9  8  7  6  5  4  3  2  1  0  
10  9  8  7  6  5  4  3  2  1  0  
10  9  8  7  6  5  4  3  2  1  0  
10  9  8  7  6  5  4  3  2  1  0  

When it should be:

10  
10  9  8  
10  9  8  7  6  
10  9  8  7  6  5  4  
10  9  8  7  6  5  4  3  2  
10  9  8  7  6  5  4  3  2  1  0   

Can someone explain to me what is happening here? And perhaps suggest a work around for such a situation.

Answer 1

You are adding the exact same object test to the ArrayList over and over again. This is why they are all getting modified together.

Everytime you want to add it, you need to create a new ArrayList<Integer>() before you add it.

You can try replacing

test1.add(test);

with

test = new ArrayList<Integer>(test);
test1.add(test);

This makes a copy of the current test and then adds it to the list.

This would still mean that the Integer elements contained in the ArrayList would exhibit the same bug, since this is a shallow copy, but since you are not modifying them, it is ok for this case.