正向循环或反向循环取决于生成的随机数？

我正在努力工作这个逻辑，但似乎无法实现。我会很感激这方面的任何方向。我想要一种方法来生成1到10之间的随机数，然后返回一个本地String []数组“numArr”的四个元素的数组，该数组是本方法的本地成员。因此，可以说，如果产生的随机数是“8”，我将不得不for循环开始的“8”反向并返回：正向循环或反向循环取决于生成的随机数？

“八” “七” “六个一” “十二五”

我有一个设置为4的局部变量“len”，“len”变量决定了需要返回多少个“numArr”元素。我确保生成的随机数小于（numArr.length - len），因此如果生成随机数为8，我的逻辑将确定并使用从8开始的反向循环。因为如果调用转发循环，它只会执行两次，因为“numArr”只有10个元素。

我正在反向for循环：

，如果目前的随机数>（numArr.length - LEN）。所以如果随机数是7,8,9或10，逻辑将循环运行。

我目前没有返回任何东西，我只是想确保在我返回任何东西之前逻辑功能完全正常。

public static void geNums(){ 

    //LENGTH IS SET TO 4 
    int len = 4; 

    //COUNTER 
    int counter = 0; 

    //LOCAL ARRAY WITH JUS 10 ELEMENTS 
    String[] numArr = {"one","two","three","four","five","six","seven","eight","nine","ten"};  

    //RANDOM NUMBER IS GENERATED BETWEEN 1 AND 10 
    Random randNum = new Random();  
    int curRandom = randNum.nextInt(9) + 0; //0 TO 9 || 1 TO 10 

    //CHOICE ARRAY 
    String[] choices = new String[ len ]; //LENGTH IS SET TO 4 

    //DISPLAY CURRENTLY GENERATED RANDOM NUMBER 
    System.out.println("Current Random Number: " + curRandom); 
    System.out.println("-------------------------"); 

    if(curRandom > (numArr.length - len)){ 

     //REVERSE LOOP 
     for(int i = curRandom; i >= len; i--){ 

     //BREAK IF COUNTER IS MORE THAN LEN 
     if(counter > len){ 
      --counter; 
      break; 
     } 
     choices[i] = numArr[i];  //POPULATE CHOICE ARRAY 
     ++counter; 
    } 
    else{ 
     //FORWARD LOOP 
     for(int i = curRandom; i < len; i++){ 
      choices[i] = numArr[i]; //POPULATE CHOICE ARRAY 
     }  
    } 

    //DISPLAY CHOICE ARRAY ELEMENTS FOR DUBUGGING 
    for(int j = 0; j < choices.length; j++){ 
     System.out.println(choices[j]); 
    } 

}//METHOD

来源

2014-02-09 Combustion007

根据我的理解，您生成一个介于1..10之间的数字。如果它大于五，则将其字符串名称保存到数组中。我已经修改了你的代码，希望你能理解。我发现的第一个问题是（int i = curRandom; i> = len; i--）...... choices[i] = numArr [i]。记住你的数组（选择）被声明为4的长度：String[] choices = new String[ len ]; //LENGTH IS SET TO 4那么如果你生成8号呢？您可以从8开始蒙山i，当你调用choices[i]等于蒙山choices[8] 它会给一个IndexOutOfBoundException，因为你的阵列只有4所以这里是一个解决方案：

//LENGTH IS SET TO 4 
      int len = 4; 

      //COUNTER 
      int counter = 0; 

      //LOCAL ARRAY WITH JUS 10 ELEMENTS 
      String[] numArr = {"one","two","three","four","five","six","seven","eight","nine","ten"};  

      //RANDOM NUMBER IS GENERATED BETWEEN 1 AND 10 
      Random randNum = new Random();  
      int curRandom = randNum.nextInt(9) + 0; //0 TO 9 || 1 TO 10 

      //CHOICE ARRAY 
      String[] choices = new String[ len ]; //LENGTH IS SET TO 4 

      //DISPLAY CURRENTLY GENERATED RANDOM NUMBER 
      System.out.println("Current Random Number: " + curRandom); 
      System.out.println("-------------------------"); 

      if(curRandom > 5){ 
       int aux = curRandom; 
       for(int i = 0;i<len;i++) { 
        choices[i] = numArr[aux]; 
        aux--; 
       } 
      } 
      else { 
       int aux = curRandom; 
       for(int i = 0;i<len;i++) { 
        choices[i] = numArr[aux]; 
        aux++; 
       } 
      } 


      //DISPLAY CHOICE ARRAY ELEMENTS FOR DUBUGGING 
      for(int j = 0; j < choices.length; j++) 
       System.out.println(choices[j]);

输出：

Current Random Number: 2 
------------------------- 
three 
four 
five 
six

来源

2014-02-09 22:32:48 user3043278

正向循环或反向循环取决于生成的随机数？

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