如何按字母顺序排列双维表格？

huangapple
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英文:

How can I put alphabetically a double dimensional table?

问题

以下是翻译好的代码部分：

  1. for (int i = 0; i < table2.length; i++) {
  2.     for (int j = 0; j < table2[0].length; j++) {
  3.         if (i == 0 && ((int) table2[i][i] > (int) table2[i][j])) {
  4.             char d = table2[i][j];
  5.             table2[i][j] = table2[i][i];
  6.             table2[i][i] = (char) d;
  7.         }
  9.         System.out.print((char) table2[i][j] + " ");
  10.         if (j == userKeyWord.length() - 1)
  11.             System.out.println();
  12.     }
  13. }

注意：这只是你提供的代码的翻译部分，可能仍然需要检查和调试，以确保其正确性和完整性。另外，你提到的 userKeyWord 部分在你提供的代码片段中并未包含在内，所以需要在实际代码中补充相关部分。

英文:

I have a double dimensional table which on his first line has letters and below every letter there are numbers. I want to put the first line alphabetically but for every letter that change it's place I want the numbers below to move together.

For example I have this table:

  1. C O R T I N A
  2. 1 1 0 0 3 4 0
  3. 4 3 3 1 2 1 3
  4. 2 1 2 1 2 2 0
  5. 0 2 3 0 2 0 4
  6. 2 3 3 4 0 4 3
  7. 2

and the result must be:

  1. A C I N O R T
  2. 0 1 3 4 1 0 0
  3. 3 4 2 1 3 3 1
  4. 0 2 2 2 1 2 1
  5. 4 0 2 0 2 3 0
  6. 3 2 0 4 3 3 4
  7.   2

I have make a try here is my code but it doesn't work. In comments is the whole code I thought I did it but it's not working. Then I tried to change just the first line and I still failed. Can someone help me?

  1. 				for (int i=0; i&lt;table2.length; i++) {
  2. 				for (int j=0; j&lt;table2[0].length; j++) {
  3. 					if (i==0 &amp;&amp; ((int)table2[i][i]&gt;(int)table2[i][j])) {
  4. 						char d = table2[i][j];
  5. 						table2[i][j]=table2[i][i];
  6. 						table2[i][i]=(char)d;
  7. //							char [] help = new char [table2[j].length];
  8. //							for (int k=0; k&lt;help.length; k++) {
  9. //								help[k]=table2[i+k][j];
  10. //								//System.out.print(help[k]);
  11. //							}
  12. //
  13. //							for (int u=0; u&lt;table2.length; u++) {
  14. //								for (int p=0; p&lt;table2[0].length; p++) {
  15. //									if (u==0 &amp;&amp; (int)table2[u][u]&gt;(int)table2[u]
    ) {
  16. //										table2[u]
     = table2[u][u];
  17. //
  18. //										for (int k=0; k&lt;help.length; k++) {
  19. //											table2[u][u] = help[k];
  20. //										}
  21. //									}
  22. //								}
  23. //							}
  25. 					}
  29. 					System.out.print((char)table2[i][j] + &quot; &quot;);
  30. 					if (j==userKeyWord.length()-1)
  31. 						System.out.println();
  32. 				}
  33. 			}

I forgot to mention the userKeyWord at the end is the word the user gives for the numbers to print below. (ofc it's just a part of the code)

答案1

得分: 1

我首先通过对第一行的字母值进行排序来排序第一行的索引。然后，我使用这些索引来重新排列其他行，以反映相同的位置。为了迎合末尾的单独值，我添加了值为-1的虚拟值来填充未使用的数组位置。这使得矩阵更加均匀且更易于处理。

  1. int[][] mat = { { 'C', 'O', 'R', 'T', 'I', 'N', 'A' },
  2. 		{ 1, 1, 0, 0, 3, 4, 0 },
  3. 		{ 4, 3, 3, 1, 2, 1, 3 },
  4. 		{ 2, 1, 2, 1, 2, 2, 0 },
  5. 		{ 0, 2, 3, 0, 2, 0, 4 },
  6. 		{ 2, 3, 3, 4, 0, 4, 3 },
  7. 		{ 2,-1,-1,-1,-1,-1,-1 } };

获取第一行并基于其进行索引排序。

  1. int[] first = mat[0];
  3. int[] indices = IntStream.range(0, first.length).boxed()
  4. 		.sorted((a, b) -> Integer.compare(mat[0][a],
  5. 				mat[0][b]))
  6. 		.mapToInt(Integer::intValue).toArray();

现在使用这些索引来构建新矩阵。

  1. for (int row = 0; row < mat.length; row++) {
  2. 	int[] temp = mat[row];
  3. 	mat[row] = Arrays.stream(indices).map(i -> temp[i])
  4. 			.toArray();
  5. }

然后打印它。

  1. for (int i = 0; i < mat[0].length; i++) {
  2. 	System.out.printf(" %c ", mat[0][i]);
  3. }
  4. System.out.println();
  6. for (int row = 1; row < mat.length; row++) {
  7. 	for (int col = 0; col < mat[row].length; col++) {
  8. 		int v = mat[row][col];
  9. 		System.out.printf("%2s ", v < 0 ? "" : v);
  10. 	}
  11. 	System.out.println();
  12. }

打印结果为：

  1.  A  C  I  N  O  R  T 
  2.  0  1  3  4  1  0  0 
  3.  3  4  2  1  3  3  1 
  4.  0  2  2  2  1  2  1 
  5.  4  0  2  0  2  3  0 
  6.  3  2  0  4  3  3  4 
  7.     2
英文:

I did this by first sorting the indices of the first row based on the letter value. Then I used those indices to rearrange the other rows to reflect the same position. To cater to the lone value at the end I added dummy values of -1 to fill the unused array locations. This makes the matrix more uniform and easier to work with.

  1. int[][] mat = { { &#39;C&#39;, &#39;O&#39;, &#39;R&#39;, &#39;T&#39;, &#39;I&#39;, &#39;N&#39;, &#39;A&#39; },
  2. 		{ 1, 1, 0, 0, 3, 4, 0 },
  3. 		{ 4, 3, 3, 1, 2, 1, 3 },
  4. 		{ 2, 1, 2, 1, 2, 2, 0 },
  5. 		{ 0, 2, 3, 0, 2, 0, 4 },
  6. 		{ 2, 3, 3, 4, 0, 4, 3 },
  7. 		{ 2,-1,-1,-1,-1,-1,-1 } };

Get the first row and sort the indices based on that.

  1. int[] first = mat[0];
  3. int[] indices = IntStream.range(0, first.length).boxed()
  4. 		.sorted((a, b) -&gt; Integer.compare(mat[0][a],
  5. 				mat[0][b]))
  6. 		.mapToInt(Integer::intValue).toArray();

Now use those indices to build a new matrix

  1. for (int row = 0; row &lt; mat.length; row++) {
  2. 	int[] temp = mat[row];
  3. 	mat[row] = Arrays.stream(indices).map(i -&gt; temp[i])
  4. 			.toArray();
  5. }

And then print it

  2. for (int i = 0; i &lt; mat[0].length; i++) {
  3. 	System.out.printf(&quot; %c &quot;, mat[0][i]);
  4. }
  5. System.out.println();
  7. for (int row = 1; row &lt; mat.length; row++) {
  8. 	for (int col = 0; col &lt; mat[row].length; col++) {
  9. 		int v = mat[row][col];
  10. 		System.out.printf(&quot;%2s &quot;, v &lt; 0 ? &quot;&quot; : v);
  11. 	}
  12. 	System.out.println();
  13. }

Prints

  1.  A  C  I  N  O  R  T 
  2.  0  1  3  4  1  0  0 
  3.  3  4  2  1  3  3  1 
  4.  0  2  2  2  1  2  1 
  5.  4  0  2  0  2  3  0 
  6.  3  2  0  4  3  3  4 
  7.     2                
  11. </details>
  15. # 答案2
  16. **得分**: 1
  18. 我不确定为什么你的代码不起作用，但我以稍微不同的方式来处理它。这是我的做法。
  20. ```java
  21. private static final char[][] unsortedArray = {{'C', 'O', 'R', 'T', 'I', 'N', 'A'},
  22.                                                 {1, 1, 0, 0, 3, 4, 0},
  23.                                                 {4, 3, 3, 1, 2, 1, 3},
  24.                                                 {2, 1, 2, 1, 2, 2, 0},
  25.                                                 {0, 2, 3, 0, 2, 0, 4},
  26.                                                 {2, 3, 3, 4, 0, 4, 3}};
  28. // 保存原始索引和键的对象
  29. private static class IndexTuple {
  30.     protected char key;
  31.     protected int unsortedIndex;
  33.     public IndexTuple(char key, int unsortedIndex) {
  34.         this.key = key;
  35.         this.unsortedIndex = unsortedIndex;
  36.     }
  37. }
  39. // 比较器，用于让集合进行排序
  40. private static class TupleComparator implements Comparator<IndexTuple> {
  42.     @Override
  43.     public int compare(IndexTuple o1, IndexTuple o2) {
  44.         return o1.key - o2.key;
  45.     }
  46. }
  48. public static char[][] sortTable(char[][] unsortedTable) {
  49.     // 创建新数组以复制结果
  50.     char[][] sortedTable = new char[unsortedTable.length][unsortedTable[0].length];
  52.     // 将结果复制到自定义对象的列表中
  53.     ArrayList<IndexTuple> tupleList = new ArrayList<>();
  54.     for (int i = 0; i < unsortedTable[0].length; i++) {
  55.         tupleList.add(new IndexTuple(unsortedTable[0][i], i));
  56.     }
  58.     // 排序列表
  59.     tupleList.sort(new TupleComparator());
  61.     // 使用原始未排序索引和新排序索引，将所有值设置为正确的顺序
  62.     for (int i = 0; i < unsortedTable.length; i++) {
  63.         for (int j = 0; j < tupleList.size(); j++) {
  64.             sortedTable[i][j] = unsortedTable[i][tupleList.get(j).unsortedIndex];
  65.         }
  66.     }
  67.     return sortedTable;
  68. }
英文:

I am not 100% sure why your code isn't working, but I approached it a little bit differently. Here is what I did.

  1. private static final char[][] unsortedArray = {{&#39;C&#39;, &#39;O&#39;, &#39;R&#39;, &#39;T&#39;, &#39;I&#39;, &#39;N&#39;, &#39;A&#39;},
  2.                                                 {1, 1, 0, 0, 3, 4, 0},
  3.                                                 {4, 3, 3, 1, 2, 1, 3},
  4.                                                 {2, 1, 2, 1, 2, 2, 0},
  5.                                                 {0, 2, 3, 0, 2, 0, 4},
  6.                                                 {2, 3, 3, 4, 0, 4, 3}};
  8. //Object to hold original index and key
  9. private static class IndexTuple {
  10.     protected char key;
  11.     protected int unsortedIndex;
  13.     public IndexTuple(char key, int unsortedIndex) {
  14.         this.key = key;
  15.         this.unsortedIndex = unsortedIndex;
  16.     }
  17. }
  19. //Comparator to let collections sort for you
  20. private static class TupleComparator implements Comparator&lt;IndexTuple&gt; {
  22.     @Override
  23.     public int compare(IndexTuple o1, IndexTuple o2) {
  24.         return o1.key - o2.key;
  25.     }
  26. }
  28. public static char[][] sortTable(char[][] unsortedTable) {
  29.     //Create new array to copy result to
  30.     char[][] sortedTable = new char[unsortedTable.length][unsortedTable[0].length];
  32.     //Copy result to List of custom objects
  33.     ArrayList&lt;IndexTuple&gt; tupleList = new ArrayList&lt;&gt;();
  34.     for (int i = 0; i &lt; unsortedTable[0].length; i++) {
  35.         tupleList.add(new IndexTuple(unsortedTable[0][i], i));
  36.     }
  38.     //sort list
  39.     tupleList.sort(new TupleComparator());
  41.     //Using original unsorted index and new sorted index set all the values in correct order
  42.     for (int i = 0; i &lt; unsortedTable.length; i++) {
  43.         for (int j = 0; j &lt; tupleList.size(); j++) {
  44.             sortedTable[i][j] = unsortedTable[i][tupleList.get(j).unsortedIndex];
  45.         }
  47.     }
  48.     return sortedTable;
  49. }

huangapple
  • 本文由 发表于 2020年10月16日 06:30:02
  • 转载请务必保留本文链接：https://go.coder-hub.com/64380520.html
  • alphabetical
  • arrays
  • java
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