将代码行从 C++ 转换为 JAVA

Question

I have come up with the below code. Managed to resolve most of the errors except the one related to Map.

I understand that the below line of code belongs to C++. Tried a lot to convert it to JAVA since couple of days, unable to figure out a way:

Below lines of code in C++

map<Character,Integer> enc = new map<Character,Integer>();

Note:Upon changing the above syntax to HashMap/Map and after importing Java.Util, lines of code marked with 3 stars in the below code displays the following error "The type of the expression must be an array type but it resolved to Map"

1) enc[input.charAt(i)] = i; 2) int pos = enc[msg.charAt(i) - 32]; 3) int pos = enc[msg.charAt(i)];

// This function will decipher any input message

public static String ABC(String msg, String input)
        {
            // Hold the position of every character (A-Z) from encoded string 
            map<Character,Integer> enc = new map<Character,Integer>();
            for (int i = 0; i < input.length(); i++)
            {
                ***enc[input.charAt(i)] = i;***
            }

            String decipher = "";

            // This loop deciphered the message. 
            // Spaces, special characters and numbers remain same. 
            for (int i = 0; i < msg.length(); i++)
            {
                if (msg.charAt(i) >= 'a' && msg.charAt(i) <= 'z')
                {
                    ***int pos = enc[msg.charAt(i) - 32];***
                    decipher += plaintext.charAt(pos);
                }
                else if (msg.charAt(i) >= 'A' && msg.charAt(i) <= 'Z')
                {
                    ***int pos = enc[msg.charAt(i)];***
                    decipher += plaintext.charAt(pos);
                }
                else
                {
                    decipher += msg.charAt(i);
                }
            }
            return decipher;
        }

Answer 1

map<Character,Integer> enc = new map<Character,Integer>();

Map is an interface type in Java, and cannot be instantated [as anything other than an anonymous inner class]. You need to instantiate one of the types that implements Map , like TreeMap or HashMap .

//Since Java 7, <> infers the type arguments
Map<Character, Integer> enc = new HashMap<>();

 enc[input.charAt(i)] = i;

The brackets operator syntax you're using ( enc[input.charAt(i)] ) is native to C++, and is not overloadable in Java; thus, the only time these kinds of brackets are allowed in Java is when using an array.

You need to use get() and put() with java maps.

enc.put(input.charAt(i), i);
//...
int pos = enc.get(msg.charAt(i) - 32);

Answer 2

Old question, but for future reference, here's the solution:

String plaintext = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public static String ABC(String msg, String input)
    {
        // Hold the position of every character (A-Z) from encoded string
        Map<Character, Integer> enc = new HashMap<>();
        for (int i = 0; i < input.length(); i++) 
        {
        des.put(input.charAt(i), i);
        } 

        String decipher = "";

        // This loop deciphered the message. 
        // Spaces, special characters and numbers remain same. 
        for (int i = 0; i < msg.length(); i++)
        {
            if (msg.charAt(i) >= 'a' && msg.charAt(i) <= 'z')
            {
                int pos = enc.get((char)(msg.charAt(i)-32));
                decipher += plaintext.charAt(pos);
            }
            else if (msg.charAt(i) >= 'A' && msg.charAt(i) <= 'Z')
            {
                int pos = enc.get(mensaje.charAt(i));
                decipher += plaintext.charAt(pos);
            }
            else
            {
                decipher += msg.charAt(i);
            }
        }
        return decipher;
    }

Basically you have to use put when mapping, and then get when using it. Oh and casting char when substracting from it.