如何通过一个循环运行一个2D数组？ [英] How to run through a 2d array with a single loop?

问题描述

  for（int）我想知道是否可以写这个东西，但只用一个循环而不是两个？行= 0;行< matrix.length;排++）{
为（INT COL = 0; COL<矩阵[0]。长度; COL ++）{
如果（（行+ COL）％ 2 == 0）{
 System.out.print（matrix [row] [col] +，）; 
 
 sum + = matrix [row] [col]; 
 
 
 
 System.out.println（和+）; 
 
 
 
 $ b实际上只是忽略了循环体不知道为什么我包括它..如何以某种方式将结合为循环是我的问题。
 
 
 
只是保持简单，如果可能的话。谢谢！
 
 
解决方案
 
可以，虽然它是低效的：
 
 
 <$ P （int i = 0; i< matrix.length * matrix [0] .length; i ++）
 sum + = matrix [i％matrix.length] [i / matrix.length]; 
  

基本思路是将每个索引映射到一个2d数字空间的值，事实上，我们知道数组的每个行的长度（ matrix.length ）。我们可以组成一个索引，它通过 z = x + y * matrix.length 来唯一标识两个索引 matrix [x] [y] / code>。然后反过来是：

$ p $ code $ x $ z $ matrix $ length
y = z / matrix.length


这个描述是完整的，例如在 [0，matrix.length * matrix [0] .length）中的每一个 z ，所以我们可以在这里使用它。

I was wondering if I could write this very thing but with one single loop, instead of two?

for (int row = 0; row < matrix.length; row++) {
        for (int col = 0; col < matrix[0].length; col++) {
            if ((row + col) % 2 == 0) {
                System.out.print(matrix[row][col] + ", ");

                sum += matrix[row][col];
            }
        }

        System.out.println("with a sum of " + sum);
    }

Actually just ignore the body of the loop.. It's totally irrelevant, I have no idea whatsoever why I included it.. How to somehow combine the two for loops is my question.

Just keep it simple, if possible. Thank you!

解决方案

You can, though it's inefficient:

for(int i = 0 ; i < matrix.length * matrix[0].length ; i++)
     sum += matrix[i % matrix.length][i / matrix.length];

The basic idea would be to map each index to a value in a 2d-number space, using the fact that we know the length of each "row" of the array (matrix.length). We can compose a single index, that uniquely identifies two indices matrix[x][y], by z = x + y * matrix.length. The reverse of this would then be:

x = z % matrix.length
y = z / matrix.length

This depiction would be complete, e.g. each z in [0 , matrix.length * matrix[0].length) would identify exactly one pair of indices, thus we can use it here.

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