Question

我想生成一个从0到11的5个随机整数序列，然后按递增顺序对其进行排序。我编码如下，但logcat显示NullPointerException。

怎么可能这样解决？非常感谢!!

public class Memory extends Activity 
{     
   Integer[] NumSeq;
   private Random random1;  

   @Override
   public void onCreate(Bundle savedInstanceState) 
   {...
      random1 = new Random();
      setQup();
   }

   private void setQup() 
   {
       NumSeq[0] = random1.nextInt(12);            
       NumSeq[1] = random1.nextInt(12); 
       while (NumSeq[1] == NumSeq[0]) {NumSeq[1] = random1.nextInt(12);}
       NumSeq[2] = random1.nextInt(12); 
       while ((NumSeq[2] == NumSeq[1]) || (NumSeq[2] ==NumSeq[0])) {NumSeq[2] = random1.nextInt(12);}

      ...

       Arrays.sort(NumSeq, new Comparator<Integer>()
                {
                    @Override
                    public int compare(Integer x, Integer y)
                    {
                        return x - y;
                    }
                });        

       Button11.setText(""+NumSeq[0]);
       Button12.setText(""+NumSeq[1]);
       Button13.setText(""+NumSeq[2]);         
   }
}

logcat的：

12-12 00:01:17.875: E/AndroidRuntime(18703): FATAL EXCEPTION: main
12-12 00:01:17.875: E/AndroidRuntime(18703): java.lang.RuntimeException: Unable to start activity ComponentInfo{com.pearappx.logic.run/com.pearappx.logic.run.Memory}: java.lang.NullPointerException
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:1967)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:1992)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.ActivityThread.access$600(ActivityThread.java:127)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1158)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.os.Handler.dispatchMessage(Handler.java:99)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.os.Looper.loop(Looper.java:137)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.ActivityThread.main(ActivityThread.java:4511)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at java.lang.reflect.Method.invokeNative(Native Method)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at java.lang.reflect.Method.invoke(Method.java:511)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:980)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:747)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at dalvik.system.NativeStart.main(Native Method)
12-12 00:01:17.875: E/AndroidRuntime(18703): Caused by: java.lang.NullPointerException
12-12 00:01:17.875: E/AndroidRuntime(18703):    at com.abc.logic.run.Memory.setQup(Memory.java:111)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at com.abc.logic.run.Memory.onCreate(Memory.java:66)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.Activity.performCreate(Activity.java:4470)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1052)
12-12 00:01:17.875: E/AndroidRuntime(18703):    at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:1931)

第111行代表上面NumSeq[0] = random1.nextInt(12);

的行

解决方案：

   private List<Integer> NumS;  

   private void setQup() 
   {
// set random number from 0 to 11   
           int num0 = random1.nextInt(12);
           int num1 = random1.nextInt(12);
           while (num1 ==num0) {num1 = random1.nextInt(12);}
           int num2 = random1.nextInt(12);
           while ((num2 ==num0)||(num2 ==num1)) {num2 = random1.nextInt(12);}
           int num3 = random1.nextInt(12);
           while ((num3 ==num0)||(num3 ==num1)||(num3 ==num2)) {num3 = random1.nextInt(12);}

       NumS = new ArrayList<Integer>();
       NumS.clear();           
       NumS.add(num0);
       NumS.add(num1);
       NumS.add(num2);
       NumS.add(num3);
       Collections.sort(NumS);

       Button11.setText(""+NumS.get(0));
               ...
}

Answer 1

NumSeq永远不会创建：

Integer[] NumSeq = new Integer[12];

Answer 2

您忘记初始化Integer[] NumSeq，因此将其初始化为：

Integer[] NumSeq = new Integer[12];

并避免使用静态大小ArrayList<Integer>代替Integer Array

您可以使用Collections.sort

轻松排序ArrayList

Collections.sort(unsortedList); // in ascending Order

Answer 3

Integer[] NumSeq = new Integer[];

Answer 4

你必须初始化NumSeq：

Integer[] NumSeq = new Integer[12]

但是，您应该避免使用静态大小的数组，原因有两个：

它限制了代码的可伸缩性。
如果使用List<T>的某个子类，则可以轻松地使用Collections.sort按任意比较器或默认比较器进行排序。见Java how to sort a Linked List?。

按升序排序整数

logcat的：

解决方案：

4 个答案: