设置整型数组（int[]）的元素为null时出现错误。

huangapple
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英文:

Error when setting entry of int[] array to null

问题

I had the following idea to take an array and remove any duplicates. However, I am getting the error "error: incompatible types: cannot be converted to int" referring to the part of the code where I set temp[i] = null. Is it possible to do this? How can I fix this problem?

  1. public static int[] withoutDuplicates(int[] x) {
  2.     int[] temp = new int[x.length];
  3.     int count = 0;
  4.     for (int i = x.length - 1; i >= 0; i--) {
  5.         for (int j = i-1; j >= 0; j--) {
  6.             if (temp[i] == x[j]) {
  7.                 temp[i] = 0; // Instead of null, assign 0 to indicate removal.
  8.                 count++;
  9.             }
  10.         }
  11.     }
  12.     int size = x.length - count;
  13.     int[] a = new int[size];
  14.     int pos = 0;
  15.     for (int i = 0; i < x.length; i++) {
  16.         if (temp[i] != 0) { // Check for 0 instead of null.
  17.             a[pos] = temp[i];
  18.             pos++;
  19.         }
  20.     }
  21.     return a;
  22. }
英文:

I had the following idea to take an array and remove any duplicates. However, I am getting the error "error: incompatible types: <null> cannot be converted to int" referring to the part of the code where I set temp[i] = null. Is it possible to do this? How can I fix this problem?

  1. public static int[] withoutDuplicates(int[] x) {
  2.     int[] temp = new int[x.length];
  3.     int count = 0;
  4.     for (int i = x.length - 1; i &gt;= 0; i--) {
  5.       for (int j = i-1; j &gt;= 0; j--) {
  6.         if (temp[i] == x[j]) {
  7.           temp[i] = null;
  8.           count++;
  9.         }
  10.       }
  11.     }
  12.     int size = x.length - count;
  13.     int[] a = new int[size];
  14.     int pos = 0;
  15.     for (int i = 0; i &lt; x.length; i++) {
  16.       if (temp[i] != null) {
  17.         a[pos] = temp[i];
  18.         pos++;
  19.       }
  20.     }
  21.     return a;
  22.   }

答案1

得分: 1

你不能将 null 赋值给原始类型 int，但可以将 null 赋值给包装对象 Integer。

要移除整型数组的重复项，你可以使用类似以下的方法：

  1. myIntArray = Arrays.stream(myIntArray).distinct().toArray();
英文:

You can't assign a null to a primitive type int but you can assign a null to the wrapper object Integer.

To remove the duplicates of an int array you could use something like this:

  1. myIntArray = Arrays.stream(myIntArray).distinct().toArray();

答案2

得分: 1

流式解决方案简洁，但根据您的任务/要求，您可以仅使用临时布尔数组来标记重复位置：

  1. public static int[] withoutDuplicates(int[] x) {
  2.     boolean[] duplicated = new boolean[x.length];
  3.     int count = 0;
  4.     for (int i = x.length - 1; i > 0; i--) {
  5.         for (int j = i-1; j >= 0; j--) {
  6.             if (x[i] == x[j]) {
  7.                 duplicated[i] = true;
  8.                 count++;
  9.              }
  10.         }
  11.     }
  12.     int size = x.length - count;
  13.     int[] a = new int[size];
  14.     for (int i = 0, pos = 0; pos < size && i < x.length; i++) {
  15.         if (!duplicated[i]) {
  16.             a[pos++] = x[i];
  17.         }
  18.     }
  19.     return a;
  20. }

测试：

  1. int[] arr1 = {1, 2, 3, 4, 3, 5, 2};
  2. int[] arr2 = {1, 2, 3, 4, 0, 5, 6};
  3.         
  4. System.out.println(Arrays.toString(withoutDuplicates(arr1)));
  5. System.out.println(Arrays.toString(withoutDuplicates(arr2)));

输出：

  1. [1, 2, 3, 4, 5]
  2. [1, 2, 3, 4, 0, 5, 6]
英文:

Stream-based solution is concise but for your task/requirements you could be using just a temporary boolean array to mark positions of duplicates:

  1. public static int[] withoutDuplicates(int[] x) {
  2.     boolean[] duplicated = new boolean[x.length];
  3.     int count = 0;
  4.     for (int i = x.length - 1; i &gt; 0; i--) {
  5.         for (int j = i-1; j &gt;= 0; j--) {
  6.             if (x[i] == x[j]) {
  7.                 duplicated[i] = true;
  8.                 count++;
  9.              }
  10.         }
  11.     }
  12.     int size = x.length - count;
  13.     int[] a = new int[size];
  14.     for (int i = 0, pos = 0; pos &lt; size &amp;&amp; i &lt; x.length; i++) {
  15.         if (!duplicated[i]) {
  16.             a[pos++] = x[i];
  17.         }
  18.     }
  19.     return a;
  20. }

Test:

  1. int[] arr1 = {1, 2, 3, 4, 3, 5, 2};
  2. int[] arr2 = {1, 2, 3, 4, 0, 5, 6};
  3.         
  4. System.out.println(Arrays.toString(withoutDuplicates(arr1)));
  5. System.out.println(Arrays.toString(withoutDuplicates(arr2)));

output

  1. [1, 2, 3, 4, 5]
  2. [1, 2, 3, 4, 0, 5, 6]

huangapple
  • 本文由 发表于 2020年10月10日 04:04:37
  • 转载请务必保留本文链接：https://go.coder-hub.com/64286578.html
  • arrays
  • java
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