将字符串转换为arraylist 在java中。

How to convert a String without separator to an ArrayList<Character>.

如何将一个没有分隔符的字符串转换为ArrayList <字符> 。

My String is like this:

我的弦是这样的:

String str = "abcd..."

I know one way of doing this is converting the String to char[] first, and then convert the char [] to ArrayList <Character>.

我知道一种方法是将字符串转换为char[]，然后将char[]转换为ArrayList <字符> 。

Is there any better way to do this? like converting directly? Considering time and performance, because I am coding with a big database.

还有什么更好的办法吗?喜欢直接转换吗?考虑到时间和性能，因为我在用一个大数据库编码。

7 个解决方案

#1

You need to add it like this.

你需要像这样添加它。

String str = "abcd...";
ArrayList<Character> chars = new ArrayList<Character>();
for (char c : str.toCharArray()) {
  chars.add(c);
}

#2

use lambda expression to do this.

用lambda表达式来做这个。

String big_data = "big-data";
ArrayList<Character> chars
        = new ArrayList<>(
                 big_data.chars()
                .mapToObj(e -> (char) e)
                .collect(
                        Collectors.toList()
                )
        );

#3

If you dn not need to modify list after it created, probably the better way would be to wrap string into class implementing List<Character> interface like this:

如果您不需要在创建列表后修改列表，那么可能更好的方法是将字符串打包到类实现列表 <字符> 接口如下:

import java.util.AbstractList;
import java.util.List;

public class StringCharacterList extends AbstractList <Character>
{
    private final String string;

    public StringCharacterList (String string)
    {
        this.string = string;
    }

    @Override
    public Character get (int index)
    {
        return Character.valueOf (string.charAt (index));
    }

    @Override
    public int size ()
    {
        return string.length ();
    }
}

And then use this class like this:

然后使用这个类:

List <Character> l = new StringCharacterList ("Hello, World!");
System.out.println (l);

#4

public static void main(String[] args) {
        // TODO Auto-generated method stub
        String str = "abcd...";
        ArrayList<Character> a=new ArrayList<Character>();
        for(int i=0;i<str.length();i++)
        {
            a.add(str.charAt(i));

        }
        System.out.println(a);
    }

#5

Sorry for the Retrobump, but this is now really easy!

抱歉，这是逆行，但现在真的很简单!

You can do this easily in Java 8 using Streams! Try this:

您可以在Java 8中使用流轻松地完成这项工作!试试这个:

String string = "testingString";
List<Character> list = string.chars().mapToObj((i) -> Character.valueOf((char)i)).collect(Collectors.toList());
System.out.println(list);

You need to use mapToObj because chars() returns an IntStream.

您需要使用mapToObj，因为chars()返回一个IntStream。

#6

you can do it like this:

你可以这样做:

import java.util.ArrayList;

public class YourClass{

    public static void main(String [] args){

        ArrayList<Character> char = new ArrayList<Character>();
        String str = "abcd...";

        for (int x = 0; x < str.length(); x ++){
            char.add(str.charAt(x));
        }
    }
}

#7

There is one-line guava solution:

有一行的guava解决方案:

Lists.charactersOf("hello");

You can pass resulting ImmutableList to constructor of ArrayList but may I ask why do you need exactly ArrayList?

您可以将生成的ImmutableList传递给ArrayList的构造函数，但是我可以问您为什么需要精确的ArrayList?

#1