why assign ArrayList to new ArrayList temp

问题

I am looking at the code for Permutations problem on leetcode. For example, [1,2,3] have the following permutations: [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]. And I found there is one sentence

ArrayList<Integer> temp = new ArrayList<Integer>(l);

I have no idea why here needs to assign the "l" to "temp". And I tried current.add(l) direclty but gave me the wrong answer. Can you help me with this?

public class Solution {

    public ArrayList<ArrayList<Integer>> permute(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

        //start from an empty list
        result.add(new ArrayList<Integer>());

        for (int i = 0; i < num.length; i++) {
            //list of list in current iteration of the array num
            ArrayList<ArrayList<Integer>> current = new ArrayList<ArrayList<Integer>>();

            for (ArrayList<Integer> l : result) {
                // # of locations to insert is largest index + 1
                for (int j = 0; j < l.size()+1; j++) {
                    // + add num[i] to different locations
                    l.add(j, num[i]);
                    ArrayList<Integer> temp = new ArrayList<Integer>(l);

                    current.add(temp);

                    //System.out.println(temp);

                    // - remove num[i] add
                    l.remove(j);
                }
            }

            result = new ArrayList<ArrayList<Integer>>(current);
        }

        return result;
    }
}

回答1:

I have no idea why here needs to assign the "l" to "temp"

He's not - that would just be:

ArrayList<Integer> temp = l;

Instead, the code creates a copy of the content of the list l refers to, in a new ArrayList. That means that future changes to the list that l refers to (such as the call to l.remove(j) immediately afterwards) don't affect the new list.

As a simple stand-alone example of that, consider:

List<String> original = new ArrayList<>();
original.add("foo");
List<String> copy = new ArrayList<>(original);
System.out.println(copy.size()); // 1
original.add("bar");
System.out.println(copy.size()); // Still 1

Admittedly the code is written in a very odd manner - until the final statement, result only ever has a single element, so iterating over it is pretty pointless - but I believe that explains the single statement you were asking about.

回答2:

If you did

current.add(l);

you would be adding the same reference to the ArrayList l to current. So, if you made some changes in one of those lists, both would be modified. In order to avoid that issue, in the line

ArrayList<Integer> temp = new ArrayList<Integer>(l);

you are creating a different ArrayList but with the same content. So, they will be different objects (different references).

来源：https://stackoverflow.com/questions/24052684/why-assign-arraylist-to-new-arraylist-temp

标签

java

arraylist