简单的UDP / TCP java Hangman

Question

我有一个使用UDP / TCP连接编写简单游戏的任务。我已经在java上有一个简单的Hangman游戏。我希望游戏在2个客户端之间或服务器客户端之间播放。这是我的刽子手代码

package gameexample;
import java.util.Arrays;
import java.util.Scanner;

public class Hangman{
public static void main(String[] args) {
    String[] words = {"writer", "that", "program"};
    // Pick random index of words array
    int randomWordNumber = (int) (Math.random() * words.length);
    // Create an array to store already entered letters
    char[] enteredLetters = new char[words[randomWordNumber].length()];
    int triesCount = 0;
    boolean wordIsGuessed = false;
    do {
    // infinitely iterate through cycle as long as enterLetter returns true
    // if enterLetter returns false that means user guessed all the letters
    // in the word e. g. no asterisks were printed by printWord
    switch (enterLetter(words[randomWordNumber], enteredLetters)) {
        case 0:
            triesCount++;
            break;
        case 1:
            triesCount++;
            break;
        case 2:
            break;
        case 3:
            wordIsGuessed = true;
            break;
    }
    } while (! wordIsGuessed);
    System.out.println("\nThe word is " + words[randomWordNumber] +
        " You missed " + (triesCount -findEmptyPosition(enteredLetters)) +
        " time(s)");
}
public static int enterLetter(String word, char[] enteredLetters)    {
    System.out.print("(Guess) Enter a letter in word ");
    // If-clause is true if no asterisks were printed so
    // word is successfully guessed
    if (! printWord(word, enteredLetters))
        return 3;
    System.out.print(" > ");
    Scanner input = new Scanner(System.in);
    int emptyPosition = findEmptyPosition(enteredLetters);
    char userInput = input.nextLine().charAt(0);
    if (inEnteredLetters(userInput, enteredLetters)) {
        System.out.println(userInput + " is already in the word");
        return 2;
    }
    else if (word.contains(String.valueOf(userInput))) {
        enteredLetters[emptyPosition] = userInput;
        return 1;
    }
    else {
        System.out.println(userInput + " is not in the word");
        return 0;
        }
}

asterisks were printed, otherwise return false */
public static boolean printWord(String word, char[] enteredLetters) {
    // Iterate through all letters in word
    boolean asteriskPrinted = false;
    for (int i = 0; i < word.length(); i++) {
        char letter = word.charAt(i);
        // Check if letter already have been entered bu user before
        if (inEnteredLetters(letter, enteredLetters))
            System.out.print(letter); // If yes - print it
        else {
            System.out.print('*');
            asteriskPrinted = true;
        }
    }
    return asteriskPrinted;
}

public static boolean inEnteredLetters(char letter, char[] enteredLetters) {
    return new String(enteredLetters).contains(String.valueOf(letter));
}


public static int findEmptyPosition(char[] enteredLetters) {
    int i = 0;
    while (enteredLetters[i] != '\u0000') i++;
    return i;
 }
}

任何人都可以告诉我如何将此代码写入客户端服务器程序，每个猜测有2个玩家等待转弯吗？当其中一人猜到最后一个字母时，游戏就结束了？