Merging two arrayLists into a new arrayList, with no duplicates and in order, in Java

问题

I am trying to "combine" two arrayLists, producing a new arrayList that contains all the numbers in the two combined arrayLists, but without any duplicate elements and they should be in order. I came up with this code below. I run through it and it makes sense to me, but Im not sure if I can be using < or > to compare get(i)'s in arrayLists. I am adding all the elements in array1 into the plusArray. Then I am going through the plusArray and comparing it to array2 to see if any of array2's elements exist inside plusArray. If they do I am doing nothing, but if they dont then I am trying to add it in its correct position. Perhaps my nested for loops being used incorrectly? Note: The ArrayLists are presorted by the user in increasing order.

     ArrayList<Integer> plusArray = new ArrayList<Integer>();
for(int i = 0; i < array1.size(); i++){
    plusArray.add(array1.get(i));
}

for(int i = 0; i < plusArray.size(); i++){
    for(int j = 0; j < array2.size(); j++){

    if(array2.get(j) < plusArray.get(i)){
        plusArray.add(i,array2.get(j));
    }
    else if(plusArray.get(i).equals(array2.get(j))){
        ;
    }
    else if(array2.get(j) > plusArray.get(i)){
        plusArray.add(i, array2.get(j));
    }

}

UPDATE: I dont get the exception below anymore. Instead it seems the program runs forever. I changed the location of where to add the elements in the < and > conditions. /// Here is the exception that I get when my array lists are: IntSet 1: { 1 2 } IntSet 2: { 1 3 4 }

Exception in thread "main" java.lang.OutOfMemoryError: Java heap space
at java.util.Arrays.copyOf(Unknown Source)
at java.util.Arrays.copyOf(Unknown Source)
at java.util.ArrayList.grow(Unknown Source)
at java.util.ArrayList.ensureCapacityInternal(Unknown Source)
at java.util.ArrayList.add(Unknown Source)
at IntSet.plus(IntSet.java:92)
at IntSetDriver.main(IntSetDriver.java:61)

回答1:

Instead of the code you wrote, you may use ArrayList.addAll() to merge the lists, Collections.sort() to sort it and finally traverse of the resulting ArrayList to remove duplicates. The aggregate complexity is thus O(n)+O(n*log(n))+O(n) which is equivalent to O(n*log(n)).

回答2:

Firstly remove duplicates:

arrayList1.removeAll(arrayList2);

Then merge two arrayList:

arrayList1.addAll(arrayList2);

Lastly, sort your arrayList if you wish:

collections.sort(arrayList1);

In case you don't want to make any changes on the existing list, first create their backup lists:

arrayList1Backup = new ArrayList(arrayList1);

回答3:

Add ArrayList1, ArrayList2 and produce a Single arraylist ArrayList3. Now convert it into

Set Unique_set = new HashSet(Arraylist3);

in the unique set you will get the unique elements.
Note

ArrayList allows to duplicate values. Set doesn't allow the values to duplicate. Hope your problem solves.

回答4:

List<String> listA = new ArrayList<String>();

    listA.add("A");
    listA.add("B");

List<String> listB = new ArrayList<String>();

    listB.add("B");
    listB.add("C");

Set<String> newSet = new HashSet<String>(listA);

    newSet.addAll(listB);
List<String> newList = new ArrayList<String>(newSet);

System.out.println("New List :"+newList);

is giving you New List :[A, B, C]

回答5:

Java 8 Stream API can be used for the purpose,

ArrayList<String> list1 = new ArrayList<>();

list1.add("A");
list1.add("B");
list1.add("A");
list1.add("D");
list1.add("G");

ArrayList<String> list2 = new ArrayList<>();

list2.add("B");
list2.add("D");
list2.add("E");
list2.add("G");

List<String> noDup = Stream.concat(list1.stream(), list2.stream())
                     .distinct()
                     .collect(Collectors.toList());
noDup.forEach(System.out::println);

En passant, it shouldn't be forgetten that distinct() makes use of hashCode().

回答6:

Perhaps my nested for loops being used incorrectly?

Hint: nested loops won't work for this problem. A simple for loop won't work either.

You need to visualize the problem.

Write two ordered lists on a piece of paper, and using two fingers to point the elements of the respective lists, step through them as you do the merge in your head. Then translate your mental decision process into an algorithm and then code.

The optimal solution makes a single pass through the two lists.

回答7:

Add elements in first arraylist

ArrayList<String> firstArrayList = new ArrayList<String>();

firstArrayList.add("A");
firstArrayList.add("B");
firstArrayList.add("C");
firstArrayList.add("D");
firstArrayList.add("E");

Add elements in second arraylist

ArrayList<String> secondArrayList = new ArrayList<String>();

secondArrayList.add("B");
secondArrayList.add("D");
secondArrayList.add("F");
secondArrayList.add("G");

Add first arraylist's elements in second arraylist

secondArrayList.addAll(firstArrayList);

Assign new combine arraylist and add all elements from both arraylists

ArrayList<String> comboArrayList = new ArrayList<String>(firstArrayList);
comboArrayList.addAll(secondArrayList);

Assign new Set for remove duplicate entries from arraylist

Set<String> setList = new LinkedHashSet<String>(comboArrayList);
comboArrayList.clear();
comboArrayList.addAll(setList);

Sorting arraylist

Collections.sort(comboArrayList);

Output

 A
 B
 C
 D
 E
 F
 G

回答8:

Your second for loop should have j++ instead of i++

回答9:

I'm not sure why your current code is failing (what is the Exception you get?), but I would like to point out this approach performs O(N-squared). Consider pre-sorting your input arrays (if they are not defined to be pre-sorted) and merging the sorted arrays:

http://www.algolist.net/Algorithms/Merge/Sorted_arrays

Sorting is generally O(N logN) and the merge is O(m+n).

回答10:

Here is one solution using java 8:

Stream.of(list1, list2)
    .flatMap(Collection::stream)
    .distinct()
    // .sorted() uncomment if you want sorted list
    .collect(Collectors.toList());

回答11:

your nested for loop

 for(int j = 0; j < array2.size(); i++){

is infinite as j will always equal to zero, on the other hand, i will be increased at will in this loop. You get OutOfBoundaryException when i is larger than plusArray.size()

回答12:

You don't have to handcode this. The problem definition is precisely the behavior of Apache Commons CollectionUtils#collate. It's also overloaded for different sort orders and allowing duplicates.

回答13:

**Add elements in Final arraylist,**
**This will Help you sure**

import java.util.ArrayList;
import java.util.List;

public class NonDuplicateList {

public static void main(String[] args) {

    List<String> l1 = new ArrayList<String>();
    l1.add("1");l1.add("2");l1.add("3");l1.add("4");l1.add("5");l1.add("6");
    List<String> l2 = new ArrayList<String>();
    l2.add("1");l2.add("7");l2.add("8");l2.add("9");l2.add("10");l2.add("3");
    List<String> l3 = new ArrayList<String>();
    l3.addAll(l1);
    l3.addAll(l2);
    for (int i = 0; i < l3.size(); i++) {
        for (int j=i+1; j < l3.size(); j++) {
             if(l3.get(i) == l3.get(j)) {
                 l3.remove(j);
            }
        }
    }
    System.out.println(l3);
}

}

Output : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

回答14:

I got your point that you don't wanna use the built-in functions for merging or remove duplicates from the ArrayList. Your first code is running forever because the outer for loop condition is 'Always True'. Since you are adding elements to plusArray, so the size of the plusArray is increasing with every addition and hence 'i' is always less than it. As a result the condition never fails and the program runs forever. Tip: Try to first merge the list and then from the merged list remove the duplicate elements. :)

来源：https://stackoverflow.com/questions/9917787/merging-two-arraylists-into-a-new-arraylist-with-no-duplicates-and-in-order-in