java 如何在java中查找字符数组中是否存在元素

How to find if an element exists within a character array in java

提问by Praveen Puglia

Here is the thing. I have a character array as follows..

这是事情。我有一个字符数组如下..

char[] modes = new char[] { 'm', 'q', 'h', 'y' };

Now I want to provide the user with the option to enter a character. If it exists in the modesarray, I'll do the necessary. For that I used...

现在我想为用户提供输入字符的选项。如果它存在于modes数组中，我会做必要的。为此我用...

//to take a character as input
mode = input.next().charAt(0);
//now to check if the array contains the character
boolean ifExists = Arrays.asList(modes).contains(mode);

But strangely ifExistsreturns false.

却诡异地ifExists返回false。

1. Any Ideas where am I doing wrong?
2. If this is a bad way of doing it, please suggest a way.

1. 任何想法我在哪里做错了？
2. 如果这是一种不好的方式，请提出一种方法。

回答by NickJ

I think it's Autoboxing - the contains() method takes an object, not a primitive.

我认为这是自动装箱 - contains() 方法接受一个对象，而不是一个原始对象。

If you use Character instead of char it will work:

如果您使用 Character 而不是 char 它将起作用：

    Character[] modes = new Character[] { 'm', 'q', 'h', 'y' };

    //to take a character as input
    Character mode = "q123".charAt(0);
    //now to check if the array contains the character
    boolean ifExists = Arrays.asList(modes).contains(mode);

returns true

返回真

回答by Pat Burke

The Arrays.asList() method is returning a List of char[] and not a List of char like you are expecting. I would recommend using the Arrays.binarySort() method like so:

Arrays.asList() 方法返回一个 char[] 列表，而不是您期望的 char 列表。我建议使用 Arrays.binarySort() 方法，如下所示：

    char[] modes = new char[] { 'm', 'q', 'h', 'y' };

    char mode = 'q';

    //now to check if the array contains the character
    int index = Arrays.binarySearch(modes, mode);
    boolean ifExists = index != -1;
    System.out.print(ifExists);

回答by sunleo

I didn't find any problem with your code and try this,

我没有发现你的代码有任何问题，试试这个，

If you use this kind of Colletions then you can do lots of operations using methods available defaultly...

如果您使用这种集合，那么您可以使用默认可用的方法进行大量操作...

List<Character> l = new ArrayList<Character>();
l.add('a');
l.add('b');
l.add('c');
System.out.println(l.contains('a'));

回答by RNJ

You could just convert to a string and then run contains

您可以只转换为字符串，然后运行包含

new String(modes).contains("" + mode);

This should then return true or false for your primitive array

然后这应该为您的原始数组返回 true 或 false

回答by Jose Tepedino

You could also use String indexOf:

您还可以使用字符串 indexOf：

boolean ifExists = new String(modes).indexOf(mode) >= 0;

or

或者

boolean ifExists = "mqhy".indexOf(mode) >= 0;