blocks|key|755277|text|public+void+printGrid()
{
+++for(int+i+=+0;+i+<+20;+i%2B%2B)
+++{
++++++for(int+j+=+0;+j+<+20;+j%2B%2B)
++++++{
+++++++++System.out.printf("%255d+",+a[i][j]);
++++++}
++++++System.out.println();
+++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|755278|并替换为|unstyled|755279|public+void+replaceGrid()
{
+++for+(int+i+=+0;+i+<+20;+i%2B%2B)
+++{
++++++for+(int+j+=+0;+j+<+20;+j%2B%2B)
++++++{
+++++++++if+(a[i][j]+==+1)
++++++++++++a[i][j]+=+x;
++++++}
+++}
}|755280|你可以一口气完成这一切：|755281|public+void+printAndReplaceGrid()
{
+++for(int+i+=+0;+i+<+20;+i%2B%2B)
+++{
++++++for(int+j+=+0;+j+<+20;+j%2B%2B)
++++++{
+++++++++if+(a[i][j]+==+1)
++++++++++++a[i][j]+=+x;
+++++++++System.out.printf("%255d+",+a[i][j]);
++++++}
++++++System.out.println();
+++}
}|755282|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>public void printGrid()
{
 for(int i = 0; i &lt; 20; i++)
 {
 for(int j = 0; j &lt; 20; j++)
 {
 System.out.printf("%5d ", a[i][j]);
 }
 System.out.println();
 }
}
</code></pre>

And to replace

<pre><code>public void replaceGrid()
{
 for (int i = 0; i &lt; 20; i++)
 {
 for (int j = 0; j &lt; 20; j++)
 {
 if (a[i][j] == 1)
 a[i][j] = x;
 }
 }
}
</code></pre>

And you can do this all in one go:

<pre><code>public void printAndReplaceGrid()
{
 for(int i = 0; i &lt; 20; i++)
 {
 for(int j = 0; j &lt; 20; j++)
 {
 if (a[i][j] == 1)
 a[i][j] = x;
 System.out.printf("%5d ", a[i][j]);
 }
 System.out.println();
 }
}
</code></pre>

blocks|key|538743|text|您可以使用Utility+mettod。Arrays.deeptoString()；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|538744|+public+static+void+main(String[]+args)+{
++++int+twoD[][]+=+new+int[4][];+
++++twoD[0]+=+new+int[1];+
++++twoD[1]+=+new+int[2];+
++++twoD[2]+=+new+int[3];+
++++twoD[3]+=+new+int[4];+

++++System.out.println(Arrays.deepToString(twoD));

}|code-block|syntax|javascript|538745|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you can use the Utility mettod. Arrays.deeptoString();

<pre><code> public static void main(String[] args) {
 int twoD[][] = new int[4][]; 
 twoD[0] = new int[1]; 
 twoD[1] = new int[2]; 
 twoD[2] = new int[3]; 
 twoD[3] = new int[4]; 

 System.out.println(Arrays.deepToString(twoD));

}
</code></pre>

blocks|key|538714|text|我在另一个问题中回答了类似这样的问题|type|unstyled|depth|inlineStyleRanges|entityRanges|data|538715|public+class+Snippet+{
++++public+static+void+main(String[]+args)+{
++++++++int+[][]lst+=+new+int[10][10];

++++++++for+(int[]+arr+:+lst)+{
++++++++++++System.out.println(Arrays.toString(arr));
++++++++}
++++}

}|code-block|syntax|javascript|538716|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Something like this that i answer in another question

<pre><code>public class Snippet {
 public static void main(String[] args) {
 int [][]lst = new int[10][10];

 for (int[] arr : lst) {
 System.out.println(Arrays.toString(arr));
 }
 }

}
</code></pre>

blocks|key|755445|text|public+static+void+printTwoDimensionalArray(int[][]+a)+{
++++for+(int+i+=+0;+i+<+a.length;+i%2B%2B)+{
++++++++for+(int+j+=+0;+j+<+a[i].length;+j%2B%2B)+{
++++++++++++System.out.printf("%25d+",+a[i][j]);
++++++++}
++++++++System.out.println();
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|755446|仅适用于整型数组|unstyled|755447|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>public static void printTwoDimensionalArray(int[][] a) {
 for (int i = 0; i &lt; a.length; i++) {
 for (int j = 0; j &lt; a[i].length; j++) {
 System.out.printf("%d ", a[i][j]);
 }
 System.out.println();
 }
}
</code></pre>

just for int array

blocks|key|755243|text|您应该先按行循环，然后按列循环，结构如下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|755244|for+...row+index...
++for+...column+index...
++++print|code-block|syntax|javascript|755245|但我猜这是家庭作业，所以你自己试试吧。|755246|在for循环中交换行/列索引，这取决于您是否需要先遍历然后向下，而不是先向下然后遍历。|755247|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You should loop by rows and then columns with a structure like

<pre><code>for ...row index...
 for ...column index...
 print
</code></pre>

but I guess this is homework so just try it out yourself.

Swap the row/column index in the for loops depending on if you need to go across first and then down, vs. down first and then across.

blocks|key|755420|text|试试这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|755421|public+static+void+main+(String+[]+args)
{
+++int+[]+[]+listTwo+=+new+int+[5][5];
++++//+2+Dimensional+array+
++++int+x+=+0;
++++int+y+=+0;
++++while+(x+<+5)+{
++++++++listTwo[x][y]+=+(int)(Math.random()*10);

++++++++while+(y+<5){
++++++++++++listTwo+[x]+[y]+=+(int)(Math.random()*10);
++++++++++++System.out.print(listTwo[x][y]%2B"+%7C+");
++++++++++++y%2B%2B;++++++++++++++++
++++++++}
++++++++System.out.println("");
++++++++y=0;
++++++++x%2B%2B;
++++}
}|code-block|syntax|javascript|755422|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

How about trying this?

<pre><code>public static void main (String [] args)
{
 int [] [] listTwo = new int [5][5];
 // 2 Dimensional array 
 int x = 0;
 int y = 0;
 while (x &lt; 5) {
 listTwo[x][y] = (int)(Math.random()*10);

 while (y &lt;5){
 listTwo [x] [y] = (int)(Math.random()*10);
 System.out.print(listTwo[x][y]+" | ");
 y++; 
 }
 System.out.println("");
 y=0;
 x++;
 }
}
</code></pre>

blocks|key|538777|text|如果您知道矩阵的maxValue+(如果元素的另一次迭代不是问题，那么可以很容易地完成)，我发现下面的代码更有效和通用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|538778|++++int+numDigits+=+(int)+Math.log10(maxValue)+%2B+1;
++++if+(numDigits+<=+1)+{
++++++++numDigits+=+2;
++++}
++++StringBuffer+buf+=+new+StringBuffer();
++++for+(int+i+=+0;+i+<+matrix.length;+i%2B%2B)+{
++++++++int[]+row+=+matrix[i];
++++++++for+(int+j+=+0;+j+<+row.length;+j%2B%2B)+{
++++++++++++int+block+=+row[j];
++++++++++++buf.append(String.format("%25"+%2B+numDigits+%2B+"d",+block));
++++++++++++if+(j+>=+row.length+-+1)+{
++++++++++++++++buf.append("\n");
++++++++++++}
++++++++}
++++}
++++return+buf.toString();|code-block|syntax|javascript|538779|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you know the maxValue (can be easily done if another iteration of the elements is not an issue) of the matrix, I find the following code more effective and generic.

<pre><code> int numDigits = (int) Math.log10(maxValue) + 1;
 if (numDigits &lt;= 1) {
 numDigits = 2;
 }
 StringBuffer buf = new StringBuffer();
 for (int i = 0; i &lt; matrix.length; i++) {
 int[] row = matrix[i];
 for (int j = 0; j &lt; row.length; j++) {
 int block = row[j];
 buf.append(String.format("%" + numDigits + "d", block));
 if (j &gt;= row.length - 1) {
 buf.append("\n");
 }
 }
 }
 return buf.toString();
</code></pre>

blocks|key|538881|text|我也是一个初学者，我刚刚设法用两个嵌套的for循环破解了这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|538882|我看了这里的答案，它们对我来说有点高级，所以我想我应该分享我的答案，以帮助所有其他的新手。|538883|附注:这是一个“打地鼠”的游戏，因此这个数组被称为“moleGrid”。|538884|public+static+void+printGrid()+{
++++for+(int+i+=+0;+i+<+moleGrid.length;+i%2B%2B)+{
++++++++for+(int+j+=+0;+j+<+moleGrid[0].length;+j%2B%2B)+{
++++++++++++if+(j+==+0+%7C%7C+j+%25+(moleGrid.length+-+1)+!=+0)+{
++++++++++++++++System.out.print(moleGrid[i][j]);
++++++++++++}
++++++++++++else+{
++++++++++++++++System.out.println(moleGrid[i][j]);
++++++++++++}
+++++++++}
+++++}
}|code-block|syntax|javascript|538885|希望它能帮上忙！|538886|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I am also a beginner and I've just managed to crack this using two nested for loops. 

I looked at the answers here and tbh they're a bit advanced for me so I thought I'd share mine to help all the other newbies out there.

P.S. It's for a Whack-A-Mole game hence why the array is called 'moleGrid'.

<pre><code>public static void printGrid() {
 for (int i = 0; i &lt; moleGrid.length; i++) {
 for (int j = 0; j &lt; moleGrid[0].length; j++) {
 if (j == 0 || j % (moleGrid.length - 1) != 0) {
 System.out.print(moleGrid[i][j]);
 }
 else {
 System.out.println(moleGrid[i][j]);
 }
 }
 }
}
</code></pre>

Hope it helps!

blocks|key|538949|text|更简单的方法，使用java+5风格的for循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|538950|Integer[][]+twoDimArray+=+{{8,+9},{8,+10}};
++++++++for+(Integer[]+array:+twoDimArray){
++++++++++++System.out.print(array[0]+%2B+"+,");
++++++++++++System.out.println(array[1]);
++++++++}|code-block|syntax|javascript|538951|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

more simpler approach , use java 5 style for loop
<pre><code>Integer[][] twoDimArray = {{8, 9},{8, 10}};
 for (Integer[] array: twoDimArray){
 System.out.print(array[0] + &quot; ,&quot;);
 System.out.println(array[1]);
 }
</code></pre>

I have a [20][20] two dimensional array that I've manipulated. In a few words I am doing a turtle project with user inputting instructions like pen up = 0 and pen down = 1. When the pen is down the individual array location, for instance [3][4] is marked with a &quot;1&quot;.
The last step of my program is to print out the 20/20 array. I can't figure out how to print it and I need to replace the &quot;1&quot; with an &quot;X&quot;. The print command is actually a method inside a class that a parent program will call. I know I have to use a loop.
<pre><code>public void printGrid() {
 System.out.println...
}
</code></pre>

How to print a two dimensional array?

我有一个我已经处理过的二维数组。简而言之，我正在做一个海龟项目，用户输入的指令如pen up =0和pen down = 1。当笔向下时，单个数组位置，例如3被标记为"1“。我的程序的最后一步是打印出20/20数组。我不知道如何打印它，我需要用"X“替换"1”。print命令实际上是父程序将调用的类中的一个方法。我知道我必须使用循环。public void printGrid() {    Sys

问如何打印一个二维数组？
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回答 9

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问如何打印一个二维数组？
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