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我目前正在用java制作一个hang man游戏，我有一个方法按难度组织的单词列表，然后在另一种方法中，我有一个if语句询问用户想要玩哪个难度。我该怎么做呢？我的代码如下：

public static int word(){
 String words[] = new String[26];
    switch(diff){
    case 1:
      words[0] = "cat";
      words[1] = "dog";
      words[2] = "book";          
      words[3] = "breakfeast";          
      words[4] = "telephone";          
      words[5] = "mixture";          
      words[6] = "music";          
      words[7] = "animal";          
      words[8] = "school";          
      words[9] = "plant";          
      words[10] = "pen";          
      words[11] = "pencil";          
      words[12] = "paper";          
      words[13] = "note";          
      words[14] = "fog";          
      words[15] = "smoke";        
      words[16] = "bake";          
      words[17] = "alone";          
      words[18] = "drive";          
      words[19] = "town";          
      words[20] = "city";          
      words[21] = "sunny";          
      words[22] = "shine";          
      words[23] = "polish";          
      words[24] = "cap";          
      words[25] = "hat";

      break;
    case 2:
      words[0] = "president";
      words[1] = "exclamation";          
      words[2] = "statement";          
      words[3] = "television";          
      words[4] = "physics";          
      words[5] = "algebra";          
      words[6] = "geometry";          
      words[7] = "difficult";          
      words[8] = "extreme";          
      words[9] = "procedure";          
      words[10] = "ship";          
      words[11] = "soldier";          
      words[12] = "lunch";          
      words[13] = "hockey";          
      words[14] = "tennis";          
      words[15] = "soccer";          
      words[16] = "football";          
      words[17] = "basketball";          
      words[18] = "bias";          
      words[19] = "magazine";          
      words[20] = "computer";          
      words[21] = "internet";          
      words[22] = "allegedly";          
      words[23] = "system";          
      words[24] = "unison";          
      words[25] = "excited";         
      break;
    case 3:
      words[0] = "amalgamation";          
      words[1] = "proclomation";          
      words[2] = "establishment";          
      words[3] = "rehabilitation";          
      words[4] = "rhinoceros";          
      words[5] = "velociraptor";         
      words[6] = "declaration";         
      words[7] = "announcement";          
      words[8] = "binomial";          
      words[9] = "polynomial";          
      words[10] = "congregation";          
      words[11] = "obligation";          
      words[12] = "structure";          
      words[13] = "description";          
      words[14] = "perscription";          
      words[15] = "subscribe";          
      words[16] = "address";          
      words[17] = "township";          
      words[18] = "mischievous";          
      words[19] = "bewildered";          
      words[20] = "accusation";          
      words[21] = "designation";          
      words[22] = "disgusting";          
      words[23] = "prolonged";          
      words[24] = "restoration";          
      words[25] = "regeneration";          
  }

int i = words.length;

Random rng = new Random();     //This block of code chooses random word from array list
int choice = rng.nextInt(words.length); //Varible storing random word
String wd = words[choice];
out.println(wd);
}



public static int gameStart(){
    Scanner qwe = new Scanner(in);
    out.println("Welcome to my Hang Man game!\n");
    out.println("What difficulty would you like to play on?\t1-3");
    int diff = qwe.nextInt();

    if (diff == 1){
        //not sure what would go here
    }
    else if (diff == 2){
        //not sure what would go here
    }
    else{
        //not sure what would go here
    }


}
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2 回答 2

2
  1. 将每个难度的单词存储在单独的文本文件中。
  2. 创建一个方法，该方法String[] getWords(int difficulty)将返回给定难度从文件加载的单词列表。
  3. 不要在一种方法中混合处理程序不同方面的代码——就像你在word()方法中所做的那样。创建一个单词列表，选择随机单词并将其显示在屏幕上听起来像是 3 个不同的方面。这意味着大约 3 种方法。
于 2013-10-20T18:34:32.527 回答
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您需要该word()方法的参数，并且需要将该方法中的单词作为字符串（而不是 int）返回。其余的例程（case 语句和随机单词选择）都可以，所以我没有在下面重复。也许是这样的：

public static String word(int diff) {
    ....
    return wd;
}

然后在你有 if 块的地方，你不需要 if 块，只需将难度传递给例程。就像是：

String myWord = word(diff);
于 2013-10-20T18:39:55.767 回答