排序给定的矩阵

给定anxn矩阵。问题是要按照严格的顺序对给定的矩阵进行排序。在这里，严格顺序意味着对矩阵进行排序，以使行中的所有元素都按升序排序，并且对于行“ i”，其中1 <= i <= n-1，行“ i”的第一个元素更大等于或等于“ i-1”行的最后一个元素。

例子：

Input : mat[][] = { {5, 4, 7},
                    {1, 3, 8},
                    {2, 9, 6} }
Output : 1 2 3
         4 5 6
         7 8 9

方法：创建大小为n ^ 2的temp []数组。从第一行开始，一个一个地复制给定矩阵的元素到temp []中。排序temp []。现在，将temp []的元素一一复制到给定的矩阵中。

C++

// C++ implementation to sort the given matrix
#include 
using namespace std;
  
#define SIZE 10
  
// function to sort the given matrix
void sortMat(int mat[SIZE][SIZE], int n)
{
    // temporary matrix of size n^2
    int temp[n * n];
    int k = 0;
  
    // copy the elements of matrix one by one
    // into temp[]
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            temp[k++] = mat[i][j];
  
    // sort temp[]
    sort(temp, temp + k);
      
    // copy the elements of temp[] one by one
    // in mat[][]
    k = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            mat[i][j] = temp[k++];
}
  
// function to print the given matrix
void printMat(int mat[SIZE][SIZE], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cout << mat[i][j] << " ";
        cout << endl;
    }
}
  
// Driver program to test above
int main()
{
    int mat[SIZE][SIZE] = { { 5, 4, 7 },
                            { 1, 3, 8 },
                            { 2, 9, 6 } };
    int n = 3;
  
    cout << "Original Matrix:\n";
    printMat(mat, n);
  
    sortMat(mat, n);
  
    cout << "\nMatrix After Sorting:\n";
    printMat(mat, n);
  
    return 0;
}

Java

// Java implementation to 
// sort the given matrix
import java.io.*;
import java.util.*;
  
class GFG {
      
    static int SIZE = 10;
  
    // function to sort the given matrix
    static void sortMat(int mat[][], int n)
    {
        // temporary matrix of size n^2
        int temp[] = new int[n * n];
        int k = 0;
      
        // copy the elements of matrix 
        // one by one into temp[]
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                temp[k++] = mat[i][j];
      
        // sort temp[]
        Arrays.sort(temp);
          
        // copy the elements of temp[]
        // one by one in mat[][]
        k = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                mat[i][j] = temp[k++];
    }
      
    // function to print the given matrix
    static void printMat(int mat[][], int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                System.out.print( mat[i][j] + " ");
            System.out.println();
        }
    }
      
    // Driver program to test above
    public static void main(String args[])
    {
        int mat[][] = { { 5, 4, 7 },
                        { 1, 3, 8 },
                        { 2, 9, 6 } };
        int n = 3;
      
        System.out.println("Original Matrix:");
        printMat(mat, n);
      
        sortMat(mat, n);
      
        System.out.println("Matrix After Sorting:");
        printMat(mat, n);
      
    }
}
  
// This code is contributed by Nikita Tiwari.

Python3

# Python3 implementation to sort
# the given matrix
  
SIZE = 10
  
# Function to sort the given matrix
def sortMat(mat, n) :
      
    # Temporary matrix of size n^2
    temp = [0] * (n * n)
    k = 0
  
    # Copy the elements of matrix  
    # one by one into temp[]
    for i in range(0, n) :
          
        for j in range(0, n) :
              
            temp[k] = mat[i][j]
            k += 1
  
    # sort temp[]
    temp.sort()
      
    # copy the elements of temp[] 
    # one by one in mat[][]
    k = 0
      
    for i in range(0, n) :
          
        for j in range(0, n) :
            mat[i][j] = temp[k]
            k += 1
  
  
# Function to print the given matrix
def printMat(mat, n) :
      
    for i in range(0, n) :
          
        for j in range( 0, n ) :
              
            print(mat[i][j] , end = " ")
              
        print()
      
      
# Driver program to test above
mat = [ [ 5, 4, 7 ],
        [ 1, 3, 8 ],
        [ 2, 9, 6 ] ]
n = 3
  
print( "Original Matrix:")
printMat(mat, n)
  
sortMat(mat, n)
  
print("\nMatrix After Sorting:")
printMat(mat, n)
  
  
# This code is contributed by Nikita Tiwari.

C#

// C# implementation to
// sort the given matrix
using System;
  
class GFG {
    static int SIZE = 10;
  
    // function to sort the given matrix
    static void sortMat(int[, ] mat, int n)
    {
        // temporary matrix of size n^2
        int[] temp = new int[n * n];
        int k = 0;
  
        // copy the elements of matrix
        // one by one into temp[]
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                temp[k++] = mat[i, j];
  
        // sort temp[]
        Array.Sort(temp);
  
        // copy the elements of temp[]
        // one by one in mat[][]
        k = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
            mat[i, j] = temp[k++];
    }
  
    // function to print the given matrix
    static void printMat(int[, ] mat, int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
            Console.Write(mat[i, j] + " ");
            Console.WriteLine();
        }
    }
  
    // Driver code
    public static void Main()
    {
        int[, ] mat = { { 5, 4, 7 },
                        { 1, 3, 8 },
                        { 2, 9, 6 } };
        int n = 3;
  
        Console.WriteLine("Original Matrix:");
        printMat(mat, n);
  
        sortMat(mat, n);
  
        Console.WriteLine("Matrix After Sorting:");
        printMat(mat, n);
    }
}
  
// This code is contributed by Sam007

输出 ：

Original Matrix:
5 4 7
1 3 8
2 9 6

Matrix After Sorting:
1 2 3
4 5 6
7 8 9

时间复杂度：O(n ² log ₂ n)。
辅助空间：O(n ² )。