blocks|key|4548931|text|在找到r.search2D(table,+5)之后，您不会重置rowcount和columncount的值。因此，对于r.search2D(table,+8)，您得到的答案也是1,1。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4548932|要解决此问题，请将函数修改为：|4548933|public+Pair+search2D(int[][]+data,+int+element,+int+rowcount,+int+columncount){}|code-block|syntax|javascript|4548934|并以search2D(data,element,0,0);的身份调用|4548935|entityMap^0|3|K|V|8|14|B|1N|K|0|0|0|2|R|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|P|$]]

After you find <code>r.search2D(table, 5)</code> you are not resetting the values of <code>rowcount</code> and <code>columncount</code>. Hence for <code>r.search2D(table, 8)</code> also you are getting answer as 1,1.

To overcome this problem modify the function as:

<pre><code>public Pair search2D(int[][] data, int element, int rowcount, int columncount){}
</code></pre>

and call as <code>search2D(data,element,0,0);</code>

blocks|key|2334563|text|这里有一个我认为应该能解决问题的版本。通常，更好的设计是不依赖全局变量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2334564|public+Pair+search2D(int[][]+data,+int+element)+{
++++return+search2D(0,+0,+data,+element);
}

public+Pair+search2D(int+i,+int+j,+int[][]+data,+int+element)+{
++++if+(i+==+data.length)+{
++++++++return+new+Pair(-1,+-1);
++++}
++++if+(j+==+data[i].length)+{
++++++++return+search2D(i+%2B+1,+0,+data,+element);
++++}
++++if+(data[i][j]+==+element)+{
++++++++return+new+Pair(i,+j);
++++}
++++return+search2D(i,+j+%2B+1,+data,+element);
}|code-block|syntax|javascript|2334565|我现在知道这可能不是你需要的帮助。你需要帮助来让你的代码正常工作。但是使用它作为替代实现，它展示了递归的一些功能。:)|2334566|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here is a version that I think should do the trick. It is normally better design not to rely on global variables. 

<pre><code>public Pair search2D(int[][] data, int element) {
 return search2D(0, 0, data, element);
}

public Pair search2D(int i, int j, int[][] data, int element) {
 if (i == data.length) {
 return new Pair(-1, -1);
 }
 if (j == data[i].length) {
 return search2D(i + 1, 0, data, element);
 }
 if (data[i][j] == element) {
 return new Pair(i, j);
 }
 return search2D(i, j + 1, data, element);
}
</code></pre>

Update: I see now that this might not be the help you needed. You wanted help to get your code to work. But use this as alternative implementation, it shows some of the powers of recursion. :)

blocks|key|337966|text|我没有考虑它的递归部分，但是rowCount和columnCount在第一次搜索后不会被重置，你应该改变你的方法，使它们在一开始都是0。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|337967|public+Pair+search2D(int[][]+data,+int+element)+{
++++return+search2D(data,+element,+0,+0);
}

public+Pair+search2D(int[][]+data,+int+element,+int+rowCount,+int+columnCount)+{
++++//your+method
}|code-block|syntax|javascript|337968|而且，没有递归的方法可能会更容易。|337969|public+Pair+search2D(int[][]+data,+int+element)+{
++++for(int+row+=+0;+row+<+int.length;+row%2B%2B)
++++++++for(int+col+=+0;+col+<+int[].length;+col%2B%2B)
++++++++++++if(data[row][col]+==+element)+return+new+Pair(row,+col);
++++return+new+Pair(-1,+-1);
}|337970|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I wasn't considering the recursion part of it but rowCount and columnCount are not reset after your first search and you should change your method so that they are both 0 at the beginning.

<pre><code>public Pair search2D(int[][] data, int element) {
 return search2D(data, element, 0, 0);
}

public Pair search2D(int[][] data, int element, int rowCount, int columnCount) {
 //your method
}
</code></pre>

Also, the way without recursion may be easier.

<pre><code>public Pair search2D(int[][] data, int element) {
 for(int row = 0; row &lt; int.length; row++)
 for(int col = 0; col &lt; int[].length; col++)
 if(data[row][col] == element) return new Pair(row, col);
 return new Pair(-1, -1);
}
</code></pre>

The code I have written for this method so far is:

<pre><code>int rowCount = 0;
int columnCount = 0;
Pair p = new Pair(0, 0);
public Pair search2D(int[][] data, int element) {
 if(data[rowCount].length==columnCount)
 {
 rowCount++;
 columnCount=0;
 }

 if(data.length &gt; rowCount)
 {
 if(data[rowCount][columnCount] == element)
 {
 p = new Pair(rowCount, columnCount);
 }
 else
 {
 columnCount++;
 search2D(data, element);
 }
 }
 return p;
}
</code></pre>

"Pair" is a class I wrote because Java only allows me to return a single number, and I'm trying to return the indices that hold the location of the element. 

In my main method, I have

<pre><code> int[][] table = new int[][] { {3, 2, 8}, {3, 5, 6} };
 System.out.println(r.search2D(table, 5));
 System.out.println(r.search2D(table, 8));
</code></pre>

However, both of the outputs are (1, 1). I'm told not to use any loops, could anyone point me in the right direction or tell me where the issue is?

Returning the location of an element in 2D array recursively

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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到目前为止，我为这个方法编写的代码是：int rowCount = 0;int columnCount = 0;Pair p = new Pair(0, 0);public Pair search2D(int[][] data, int element) {     if(data[rowCount].length==...

问递归返回元素在二维数组中的位置
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问递归返回元素在二维数组中的位置EN