找到一种在空白位置填充1和0的矩阵的方法

给定一个N * M矩阵mat [] [] ，它由两种类型的字符“。”组成。和‘_’ 。任务是在包含“。”的位置填充矩阵。与1和0 。填充矩阵，以使两个相邻的单元格都不包含相同的数字，并打印修改后的矩阵。

例子：

Input: mat[][] = {{‘.’, ‘_’}, {‘_’, ‘.’}}
Output:
1 _
_ 1

Input: mat[][] = {{‘_’, ‘_’}, {‘_’, ‘_’}}
Output:
_ _
_ _
There is no place to fill the numbers.

方法：一种有效的方法是按以下模式填充矩阵：

10101010…
01010101…
10101010…

遇到任何情况都跳过“ _”字符。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define N 2
#define M 2
  
// Function to generate and
// print the required matrix
void Matrix(char a[N][M])
{
    char ch = '1';
  
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
  
            // Replace the '.'
            if (a[i][j] == '.')
                a[i][j] = ch;
  
            // Toggle number
            ch = (ch == '1') ? '0' : '1';
  
            cout << a[i][j] << " ";
        }
        cout << endl;
  
        // For each row, change
        // the starting number
        if (i % 2 == 0)
            ch = '0';
        else
            ch = '1';
    }
}
  
// Driver code
int main()
{
    char a[N][M] = { { '.', '_' },
                     { '_', '.' } };
  
    Matrix(a);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
  
class GFG 
{
  
static int N = 2;
static int M = 2;
  
// Function to generate and
// print the required matrix
static void Matrix(char a[][])
{
    char ch = '1';
  
    for (int i = 0; i < N; i++) 
    {
        for (int j = 0; j < M; j++) 
        {
  
            // Replace the '.'
            if (a[i][j] == '.')
                a[i][j] = ch;
  
            // Toggle number
            ch = (ch == '1') ? '0' : '1';
  
            System.out.print( a[i][j] + " ");
        }
        System.out.println();
  
        // For each row, change
        // the starting number
        if (i % 2 == 0)
            ch = '0';
        else
            ch = '1';
    }
}
  
    // Driver code
    public static void main (String[] args) 
    {
        char a[][] = { { '.', '_' },
                    { '_', '.' } };
  
        Matrix(a);
    }
}
  
// This code is contributed by anuj_67..

Python3

# Python3 implementation of the approach 
  
N = 2
M = 2
  
# Function to generate and 
# print the required matrix 
def Matrix(a) : 
    ch = '1'; 
  
    for i in range(N) :
        for j in range(M) :
  
            # Replace the '.' 
            if (a[i][j] == '.') :
                a[i][j] = ch; 
  
            # Toggle number 
            if (ch == '1') :
                ch == '0'
            else :
                ch = '1'
  
            print(a[i][j],end = " "); 
              
        print() 
  
        # For each row, change 
        # the starting number 
        if (i % 2 == 0) :
            ch = '0'; 
        else :
            ch = '1'; 
  
# Driver code 
if __name__ == "__main__" : 
  
    a = [
            [ '.', '_' ], 
            [ '_', '.' ],
        ]
  
    Matrix(a); 
      
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
      
class GFG 
{
  
static int N = 2;
static int M = 2;
  
// Function to generate and
// print the required matrix
static void Matrix(char [,]a)
{
    char ch = '1';
  
    for (int i = 0; i < N; i++) 
    {
        for (int j = 0; j < M; j++) 
        {
  
            // Replace the '.'
            if (a[i,j] == '.')
                a[i,j] = ch;
  
            // Toggle number
            ch = (ch == '1') ? '0' : '1';
  
            Console.Write( a[i,j] + " ");
        }
        Console.WriteLine();
  
        // For each row, change
        // the starting number
        if (i % 2 == 0)
            ch = '0';
        else
            ch = '1';
    }
}
  
// Driver code
public static void Main (String[] args) 
{
    char [,]a = { { '.', '_' },
                { '_', '.' } };
  
    Matrix(a);
}
}
  
// This code has been contributed by 29AjayKumar

输出：

1 _ 
_ 1