如何按字母顺序排列双维表格?

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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:

for (int i = 0; i < table2.length; i++) {
    for (int j = 0; j < table2[0].length; j++) {
        if (i == 0 && ((int) table2[i][i] > (int) table2[i][j])) {
            char d = table2[i][j];
            table2[i][j] = table2[i][i];
            table2[i][i] = (char) d;
        }

        System.out.print((char) table2[i][j] + " ");
        if (j == userKeyWord.length() - 1)
            System.out.println();
    }
}

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:

C O R T I N A
1 1 0 0 3 4 0
4 3 3 1 2 1 3
2 1 2 1 2 2 0
0 2 3 0 2 0 4
2 3 3 4 0 4 3
2 

and the result must be:

A C I N O R T
0 1 3 4 1 0 0
3 4 2 1 3 3 1
0 2 2 2 1 2 1
4 0 2 0 2 3 0
3 2 0 4 3 3 4
  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?

				for (int i=0; i&lt;table2.length; i++) {
				for (int j=0; j&lt;table2[0].length; j++) {
					if (i==0 &amp;&amp; ((int)table2[i][i]&gt;(int)table2[i][j])) {
						char d = table2[i][j];
						table2[i][j]=table2[i][i];
						table2[i][i]=(char)d;
//							char [] help = new char [table2[j].length];
//							for (int k=0; k&lt;help.length; k++) {
//								help[k]=table2[i+k][j];
//								//System.out.print(help[k]);
//							}
//
//							for (int u=0; u&lt;table2.length; u++) {
//								for (int p=0; p&lt;table2[0].length; p++) {
//									if (u==0 &amp;&amp; (int)table2[u][u]&gt;(int)table2[u]

) { // table2[u]

= table2[u][u]; // // for (int k=0; k&lt;help.length; k++) { // table2[u][u] = help[k]; // } // } // } // } } System.out.print((char)table2[i][j] + &quot; &quot;); if (j==userKeyWord.length()-1) System.out.println(); } }

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。

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

获取第一行并基于其进行索引排序。

int[] first = mat[0];

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

现在使用这些索引来构建新矩阵。

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

然后打印它。

for (int i = 0; i < mat[0].length; i++) {
	System.out.printf(" %c ", mat[0][i]);
}
System.out.println();

for (int row = 1; row < mat.length; row++) {
	for (int col = 0; col < mat[row].length; col++) {
		int v = mat[row][col];
		System.out.printf("%2s ", v < 0 ? "" : v);
	}
	System.out.println();
}

打印结果为:

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

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

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

int[] first = mat[0];

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

Now use those indices to build a new matrix

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

And then print it


for (int i = 0; i &lt; mat[0].length; i++) {
	System.out.printf(&quot; %c &quot;, mat[0][i]);
}
System.out.println();

for (int row = 1; row &lt; mat.length; row++) {
	for (int col = 0; col &lt; mat[row].length; col++) {
		int v = mat[row][col];
		System.out.printf(&quot;%2s &quot;, v &lt; 0 ? &quot;&quot; : v);
	}
	System.out.println();
}

Prints

 A  C  I  N  O  R  T 
 0  1  3  4  1  0  0 
 3  4  2  1  3  3  1 
 0  2  2  2  1  2  1 
 4  0  2  0  2  3  0 
 3  2  0  4  3  3  4 
    2                



</details>



# 答案2
**得分**: 1

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。

```java
private static final char[][] unsortedArray = {{'C', 'O', 'R', 'T', 'I', 'N', 'A'},
                                                {1, 1, 0, 0, 3, 4, 0},
                                                {4, 3, 3, 1, 2, 1, 3},
                                                {2, 1, 2, 1, 2, 2, 0},
                                                {0, 2, 3, 0, 2, 0, 4},
                                                {2, 3, 3, 4, 0, 4, 3}};

// 保存原始索引和键的对象
private static class IndexTuple {
    protected char key;
    protected int unsortedIndex;

    public IndexTuple(char key, int unsortedIndex) {
        this.key = key;
        this.unsortedIndex = unsortedIndex;
    }
}

// 比较器,用于让集合进行排序
private static class TupleComparator implements Comparator<IndexTuple> {

    @Override
    public int compare(IndexTuple o1, IndexTuple o2) {
        return o1.key - o2.key;
    }
}

public static char[][] sortTable(char[][] unsortedTable) {
    // 创建新数组以复制结果
    char[][] sortedTable = new char[unsortedTable.length][unsortedTable[0].length];

    // 将结果复制到自定义对象的列表中
    ArrayList<IndexTuple> tupleList = new ArrayList<>();
    for (int i = 0; i < unsortedTable[0].length; i++) {
        tupleList.add(new IndexTuple(unsortedTable[0][i], i));
    }

    // 排序列表
    tupleList.sort(new TupleComparator());

    // 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
    for (int i = 0; i < unsortedTable.length; i++) {
        for (int j = 0; j < tupleList.size(); j++) {
            sortedTable[i][j] = unsortedTable[i][tupleList.get(j).unsortedIndex];
        }
    }
    return sortedTable;
}
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.

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;},
                                                {1, 1, 0, 0, 3, 4, 0},
                                                {4, 3, 3, 1, 2, 1, 3},
                                                {2, 1, 2, 1, 2, 2, 0},
                                                {0, 2, 3, 0, 2, 0, 4},
                                                {2, 3, 3, 4, 0, 4, 3}};

//Object to hold original index and key
private static class IndexTuple {
    protected char key;
    protected int unsortedIndex;

    public IndexTuple(char key, int unsortedIndex) {
        this.key = key;
        this.unsortedIndex = unsortedIndex;
    }
}

//Comparator to let collections sort for you
private static class TupleComparator implements Comparator&lt;IndexTuple&gt; {

    @Override
    public int compare(IndexTuple o1, IndexTuple o2) {
        return o1.key - o2.key;
    }
}

public static char[][] sortTable(char[][] unsortedTable) {
    //Create new array to copy result to
    char[][] sortedTable = new char[unsortedTable.length][unsortedTable[0].length];

    //Copy result to List of custom objects
    ArrayList&lt;IndexTuple&gt; tupleList = new ArrayList&lt;&gt;();
    for (int i = 0; i &lt; unsortedTable[0].length; i++) {
        tupleList.add(new IndexTuple(unsortedTable[0][i], i));
    }

    //sort list
    tupleList.sort(new TupleComparator());

    //Using original unsorted index and new sorted index set all the values in correct order
    for (int i = 0; i &lt; unsortedTable.length; i++) {
        for (int j = 0; j &lt; tupleList.size(); j++) {
            sortedTable[i][j] = unsortedTable[i][tupleList.get(j).unsortedIndex];
        }

    }
    return sortedTable;
}

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
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