blocks|key|1023815|text|在行中添加其他元音字符之前，请检查它是否已经包含该元音，使用continue，您可以传递给另一个元音|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1023816|，您也可以对列执行相同的操作，只需切换它|unordered-list-item|1023817|
即可|1023818|在此之前：|1023819|board[row][pos]+=+tempVowels.get(temp);|code-block|syntax|javascript|1023820|写下这个：|1023821|+boolean+b+=+false;

++++for(int+j+=+0;j<columnLength;+j%2B%2B){
++++++++if(board[row][j]+==+tempVowels.get(temp))
++++++++++++b=+true;

++++++++if(b+==+true)
++++++++{
++++++++++++b+=+false;
++++++++++++continue;
++++++++}
++++++++board[row][pos]+=+tempVowels.get(temp);
++++}|1023822|entityMap^0|U|8|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$]]|$1|I|3|J|5|6|7|11|8|@]|D|@]|E|$]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|13|8|@]|D|@]|E|$P|Q]]|$1|R|3|S|5|6|7|14|8|@]|D|@]|E|$]]|$1|T|3|U|5|O|7|15|8|@]|D|@]|E|$P|Q]]|$1|V|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|W|$]]

<ol>
<li>Before adding an additional vowel character in a row check if it
already contains this vowel and with <code>continue</code> you can pass to the
other vowel</li>
<li>You can also do the same thing for the columns with just switching
it</li>
</ol>

before this:

<pre><code>board[row][pos] = tempVowels.get(temp);
</code></pre>

Write this:

<pre><code> boolean b = false;

 for(int j = 0;j&lt;columnLength; j++){
 if(board[row][j] == tempVowels.get(temp))
 b= true;

 if(b == true)
 {
 b = false;
 continue;
 }
 board[row][pos] = tempVowels.get(temp);
 }
</code></pre>

blocks|key|2044573|text|如果第一列/第一行的内容是A，而您位于第一列/第二行，则可以使用截断数组(即String+[]+availableVowels+=+{"E","I","O","U"};+)进行选择。如果你选择O，那么当你在第一列/第三行时，你应该选择from+String+[]+availableVowels+=+{"E","I","U"};。等。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2044574|entityMap^0|D|1|12|1A|2O|1|3E|16|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

If the content of first column / first row is <code>A</code> and you are on the first column / second row, you could use a truncated array, i.e. <code>String [] availableVowels = {"E","I","O","U"};</code>, to select from. If you pick <code>O</code>, then when you are on first column / thrid row, you would choose from <code>String [] availableVowels = {"E","I","U"};</code>. etc.

blocks|key|1582523|text|这与一个相当知名的问题有关，称为Rooks+Problem。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1582524|我可以建议一个更简单的循环吗？|1582525|编辑:在阅读评论后，我发现问题需要应用于每个元音。在我看来，这更具可读性：|1582526|java.util.Random+random+=+new+Random();

boolean[]+r_occupied;
boolean[]+c_occupied;

for+(i+=+0;+i+<+vowels.length;+i%2B%2B)
++{
++++//+Clear+the+'occupied'+information
++++r_occupied+=+new+boolean[5];
++++c_occupied+=+new+boolean[5];

++++//+we+will+put+vowel[i]+'count'+times+into+the+'board'
++++count+=+random.nextInt(5);

++++for+(j+=+0;+j+<+count;+j%2B%2B)
++++++{
++++++++//+generate+a+random+row
++++++++row+=+random.nextInt(5);

++++++++//+if+it+is+already+occupied,+select+the+next+one
++++++++while+(r_occupied[row])
++++++++++row+=+(row+%2B+1)+%25+5;

++++++++//+generate+a+random+column
++++++++col+=+random.nextInt(5);

++++++++//+if+it+is+already+occupied,+select+the+next+one
++++++++while+(c_occupied[col])
++++++++++col+=+(col+%2B+1)+%25+5;

++++++++/*+put+the+vowel+at+board[row][col]+*/
++++++++r_occupied[row]+=+true;
++++++++c_occupied[col]+=+true;
++++++++board[row][col]+=+vowel[i];
++++++}
++}|code-block|syntax|javascript|1582527|注意:它会覆盖一些元音，但这应该没问题。|1582528|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Rook_polynomial#Rooks_problems^0|G|D|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|11|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|12|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

This is related to a rather well known problem, called the <a href="http://en.wikipedia.org/wiki/Rook_polynomial#Rooks_problems" rel="nofollow">Rooks Problem</a>.

Might I suggest a simpler loop?

EDIT: After reading comments, I see that the problem needs to be applied to each vowel. In my opinion, this is more readable:

<pre><code>java.util.Random random = new Random();

boolean[] r_occupied;
boolean[] c_occupied;

for (i = 0; i &lt; vowels.length; i++)
 {
 // Clear the 'occupied' information
 r_occupied = new boolean[5];
 c_occupied = new boolean[5];

 // we will put vowel[i] 'count' times into the 'board'
 count = random.nextInt(5);

 for (j = 0; j &lt; count; j++)
 {
 // generate a random row
 row = random.nextInt(5);

 // if it is already occupied, select the next one
 while (r_occupied[row])
 row = (row + 1) % 5;

 // generate a random column
 col = random.nextInt(5);

 // if it is already occupied, select the next one
 while (c_occupied[col])
 col = (col + 1) % 5;

 /* put the vowel at board[row][col] */
 r_occupied[row] = true;
 c_occupied[col] = true;
 board[row][col] = vowel[i];
 }
 }
</code></pre>

Note: It will overwrite some vowels, but this should be OK.

I'm trying to generate a Sudoku board via a 2D array: board[5][5]. The Sudoku board should only contain unique vowels. However, I have only made unique vowels appear in a row. For the columns, they still seem to have duplicates. How am I suppose to generate a column with no duplicates using the code I have so far?

Here's the code I have for generating unique letters in a row:

<pre><code>String [] vowels = {"A","E","I","O","U"};
String [][] board = new String [vowels.length][5];

public Actions(){
 int rows = 5;
 for(int row = 0;row&lt;rows;row++){
 ArrayList&lt;String&gt; tempVowels = new ArrayList&lt;String&gt;(Arrays.asList(vowels));
 int numVowPerLine = (int)Math.floor(Math.random()*4);
 for(int j = 0;j&lt;numVowPerLine;j++){
 do{
 int pos = (int)Math.floor(Math.random()*5);
 if(board[row][pos] == null){
 int temp = (int)Math.floor(Math.random()*tempVowels.size());
 board[row][pos] = tempVowels.get(temp);
 tempVowels.remove(temp);
 break;
 } 
 }while(true);
 }

 }
</code></pre>

Credits to: L7ColWinters

How to delete the duplicate in a 2D array in java?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试通过一个二维数组生成一个数独棋盘: board5。数独棋盘应该只包含唯一的元音。但是，我只使唯一的元音出现在一行中。对于列，它们似乎仍然有重复项。使用到目前为止的代码，我应该如何生成没有重复的列？下面是我用来在一行中生成唯一字母的代码：String [] vowels = {"A","E","I","O","...

问如何在java中删除二维数组中的重复项？
EN

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问如何在java中删除二维数组中的重复项？EN