从String []转换为Arraylist，删除元素并转换并返回String []

Question

String [] TxtStr = new String[] {"bob","alan", "sam"};
List<String> stringList = new ArrayList<String>(Arrays.asList(TxtStr));

String [] TxtStr2 = new String[] {"bob","alan"};
List<String> stringList2 = new ArrayList<String>(Arrays.asList(TxtStr));

stringList.removeAll(stringList2);
String[] bTxtStr = stringList.toArray(new String[stringList.size()]);

String output = "nnn";

for (int x=0; x<bTxtStr.length; x++){
    output +=bTxtStr[x];
}

Currently this is a small segment of an android project I'm working on where I have to compare the contents of 2 String[].

I've been having quite a few problems so I've started out with a simple case of 2 String[] with 2 and 3 elements respectively. After initializing the String[] I convert them to ArrayLists and perform the removeAll function, which ideally should remove the elements "bob" and "alan" from the first list and eventually the output.

BTW using remove to stringList.remove("bob") works in terms of eliminating that particular string from the ArrayList. Just wondering what I can do to ensure that stringList.removeAll(....) works.

Thanks for any help in advance.

Answer 1

you have bug in:

List<String> stringList2 = new ArrayList<String>(Arrays.asList(TxtStr));

you should use TxtStr2 :

List<String> stringList2 = new ArrayList<String>(Arrays.asList(TxtStr2));

Answer 2

• reomoveAll method of ArrayList definitely works fine. You have made mistake while creating second ArrayList.
• For easy approach, you can use Java 8 Streams. Try the below code. It will give you directly single comma separated string. You can change separation character.

List<String> stringList = Arrays.asList("bob","alan", "sam");
List<String> stringList2 = Arrays.asList("bob","alan");
String mergedString = stringList.stream().filter(string ->!stringList2.contains(string)).collect(Collectors.joining(","));
System.out.println("Merged String: " + mergedString);

Answer 3

You might be ending up removing everything, because of your list creation which are based on same arrays. Change

List<String> stringList2 = new ArrayList<String>(Arrays.asList(TxtStr);

to

List<String> stringList2 = new ArrayList<String>(Arrays.asList(TxtStr2);