c++ - 在一行上打印矩阵

Question

我有这个矩阵：

1 2 3 4 5
6 7 8 9 A
B C D E F
0 1 2 3 4

我希望它打印在一行中，如下所示：

1 6 2 B 7 3 0 C 8 4 1 D 9 5 2 E A 3 F 4.

最简单的方法是如何做到这一点？

Answer 1

使用嵌套循环。外环距离 (0,0)

内循环 i 和 j 的所有有效组合，总和为距离。

Answer 2

#include <iostream>

int main(int argc, char **argv)
{
    char m[4][5] = { { '1', '2', '3', '4', '5' },
             { '6', '7', '8', '9', 'A' },
             { 'B', 'C', 'D', 'E', 'F' },
             { '0', '1', '2', '3', '4' } };

    for (int i = 0; i <= 3; ++i) {
        for (int j = 0; j <= i; ++j) {
            std::cout << m[i - j][j] << " ";
        }
    }

    for (int i = 4; i <= 7; ++i) {
        for (int j = i - 3; j <= 4; ++j) {
            std::cout << m[i - j][j] << " ";
        }
    }

    std::cout << std::endl;
    return 0;
}

Answer 3

如果矩阵存储在这样的数组中：

#include <stdio.h> 
int main(void) { 
  char matrix[4][5] = {{'1','2','3','4','5'},
                       {'6','7','8','9','A'},
                       {'B','C','D','E','F'},
                       {'0','1','2','3','4'}};
  int cols = 5;
  int rows = 4;
  int i = 0;
  for( i = 0; i < cols + rows -2 ; i++){
    int j = 0;
    while(j <= i){
        int k = i-j;
        if(k < rows && j < cols){
            printf("%c ",matrix[k][j]);
        }
        j++;
    }
  }
} 

对于总索引的每个值，我只是从 0 到行和列的总索引（在这种情况下从 0 到 7），打印值，列和行的总和等于当前总索引，如果索引可用（较小比列和行的索引），打印它，否则转义。例如：

0 - 00
1 - 10 01
2 - 20 11 02

顺便说一句，似乎需要比其他更多的循环

Answer 4

I would suggest a vector<vector<char> > representation for the matrix(if I assume that you store chars in each cell). An important observation is that for each diagonal you have the sum of the i and j indices a constant and this sum increases by one for the diagonals.

Having noticed this you make an outer cycle over the sum and inner cycle over the x coordinate. Beware not to fall out of the matrix! Now the easiest c++ way to print the matrix would be:

vector<vector<char> > a;
for(unsigned sum = 0;  sum < a.size() + a[0].size(); ++sum) {
  for (unsigned j = 0; j <= sum && j < a[0].size(); ++j) {
    unsigned i = sum - j;
    if (i >= a.size()) {
      continue;
    }
    cout << a[i][j] << " ";
  }
}

One can optimize the cycle over j by changing the start value(so that the continue condition is never true) but the code would have been harder to understand. Another nit I did not fix on purpose is that an blank is printed even after the last element. This is yeat another check that needs to be added.

Hope this helps.

Answer 5

for( int i = 0; i < matrix.cx; i++ ) {
    for( int j = 0; j < matrix.cy; j++ ) {
        std::cout << matrix[i][j] << ' ';
    }
    // comment following line to make matrix printed in one line
    std::cout << std::endl;
}