如何将矩阵的索引映射到一维数组（C++）？
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How to map the indexes of a matrix to a 1-dimensional array (C++)?
提问by Fernando Aires Castello
I have an 8x8 matrix, like this:
我有一个 8x8 矩阵，如下所示：
char matrix[8][8];
Also, I have an array of 64 elements, like this:
另外，我有一个包含 64 个元素的数组，如下所示：
char array[64];
Then I have drawn the matrix as a table, and filled the cells with numbers, each number being incremented from left to right, top to bottom.
然后我将矩阵绘制为表格，并用数字填充单元格，每个数字从左到右、从上到下递增。
If I have, say, indexes 3 (column) and 4 (row) into the matrix, I know that it corresponds to the element at position 35 in the array, as it can be seen in the table that I've drawn. I believe there is some sort of formula to translate the 2 indexes of the matrix into a single index of the array, but I can't figure out what it is.
如果我有，比如说，索引 3（列）和 4（行）到矩阵中，我知道它对应于数组中位置 35 的元素，正如在我绘制的表格中看到的那样。我相信有某种公式可以将矩阵的 2 个索引转换为数组的单个索引，但我不知道它是什么。
Any ideas?
有任何想法吗？
回答by DilithiumMatrix
The way most languages store multi-dimensional arrays is by doing a conversion like the following:
大多数语言存储多维数组的方式是进行如下转换：
If matrixhas size, n by m [i.e. i goes from 0 to (n-1) and j from 0 to (m-1) ], then:
如果matrix有大小，n x m [即 i 从 0 到 (n-1) 和 j 从 0 到 (m-1) ]，则：
matrix[ i ][ j ] = array[ i*m + j ].
matrix[ i ][ j ] = array[ i*m + j ].
So its just like a number system of base 'n'. Note that the size of the last dimension doesn't matter.
所以它就像一个以'n'为基数的数字系统。请注意，最后一个维度的大小无关紧要。
For a conceptual understanding, think of a (3x5) matrix with 'i' as the row number, and 'j' as the column number. If you start numbering from i,j = (0,0) --> 0. For 'row-major'ordering (like this), the layout looks like:
为了概念上的理解，可以想象一个 (3x5) 矩阵，其中 'i' 为行号，'j' 为列号。如果您从i,j = (0,0) --> 0. 对于“行优先”排序（像这样），布局如下所示：
|-------- 5 ---------|
Row ______________________ _ _
0 |0 1 2 3 4 | |
1 |5 6 7 8 9 | 3
2 |10 11 12 13 14| _|_
|______________________|
Column 0 1 2 3 4
As you move along the row (i.e. increase the column number), you just start counting up, so the Array indices are 0,1,2.... When you get to the second row, you already have 5entries, so you start with indices 1*5 + 0,1,2.... On the third row, you have 2*5entries already, thus the indices are 2*5 + 0,1,2....
当您沿着行移动（即增加列数）时，您只是开始计数，因此数组索引为0,1,2.... 当你到达第二行时，你已经有了5条目，所以你从索引开始1*5 + 0,1,2...。在第三行，您已经有了2*5条目，因此索引是2*5 + 0,1,2...。
For higher dimension, this idea generalizes, i.e. for a 3D matrixL by N by M:
对于更高维度，这个想法可以概括，即对于 3D matrixL by N by M：
matrix[ i ][ j ][ k ] = array[ i*(N*M) + j*M + k ]
matrix[ i ][ j ][ k ] = array[ i*(N*M) + j*M + k ]
and so on.
等等。
For a really good explanation, see: http://www.cplusplus.com/doc/tutorial/arrays/; or for some more technical aspects: http://en.wikipedia.org/wiki/Row-major_order
有关非常好的解释，请参阅：http: //www.cplusplus.com/doc/tutorial/arrays/；或更多技术方面：http: //en.wikipedia.org/wiki/Row-major_order
回答by mvelusce
For row-major ordering, I believe the statement matrix[ i ][ j ] = array[ i*n + j ]is wrong.
对于行优先排序，我认为该语句matrix[ i ][ j ] = array[ i*n + j ]是错误的。
The offset should be offset = (row * NUMCOLS) + column.
偏移量应该是offset = (row * NUMCOLS) + column。
Your statement results to be row * NUMROWS + column, which is wrong.
你的陈述结果是row * NUMROWS + column，这是错误的。
The links you provided give a correct explanation.
您提供的链接给出了正确的解释。
回答by Donny Verduijn
Something like this?
像这样的东西？
//columns = amount of columns, x = column, y = row
var calculateIndex = function(columns, x, y){
return y * columns + x;
};
The example below converts an index back to x and y coordinates.
下面的示例将索引转换回 x 和 y 坐标。
//i = index, x = amount of columns, y = amount of rows
var calculateCoordinates = function(index, columns, rows){
//for each row
for(var i=0; i<rows; i++){
//check if the index parameter is in the row
if(index < (columns * i) + columns && index >= columns * i){
//return x, y
return [index - columns * i, i];
}
}
return null;
};
