对矩阵的主对角线进行排序

给定矩阵mat[][] ，任务是按升序对矩阵的主对角元素进行排序。
主对角线：矩阵的主对角线或主对角线是元素 mat _{i, j}的集合，其中i == j 。
例子：

Input: mat[][] = {{4, 2}, {3, 1}}
Output: 
1 2
3 4
Explanation:
In the given matrix, the diagonal elements are -
=> {mat[0][0], mat[1][1]}
=> {4, 1}
=> Sorted Order = {1, 4}

Input: mat[][] = {{9, 4}, {3, 4}}     
Output: 
4 4
3 9
Explanation:
In the given matrix, the diagonal elements are -
=> {mat[0][0], mat[1][1]}
=> {9, 4}
=> Sorted Order = {4, 9}

方法：思路是修改选择排序，对矩阵的对角元素进行排序。矩阵 M*N 的对角线元素的计数将为min(M, N) 。我们知道矩阵的主要对角元素是 mat _{i, j}其中 i == j。因此，矩阵主对角线的^第i^个元素将是 mat[i][i]。因此，重复从矩阵的主对角线上找到最小元素并将其放在开头。
下面是上述方法的实现：

C++

// C++ implementation to sort the
// major diagonal of the matrix
 
#include 
 
using namespace std;
 
// Function to sort the major
// diagonal of the matrix
void sortDiagonal(int a[2][2], int M, int N)
{
    // Loop to find the ith minimum
    // element from the major diagonal
    for (int i = 0; i < M; i++) {
        int sm = a[i][i];
        int pos = i;
         
        // Loop to find the minimum
        // element from the unsorted matrix
        for (int j = i + 1; j < N; j++) {
            if (sm > a[j][j]) {
                sm = a[j][j];
                pos = j;
            }
        }
         
        // Swap to put the minimum
        // element at the beginning of
        // the major diagonal of matrix
        swap(a[i][i], a[pos][pos]);
    }
 
    // Loop to print the matrix
    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++)
            cout << a[i][j] << " ";
        cout << endl;
    }
}
 
// Driven Code
int main()
{
    int a[2][2] = { { 4, 2 },
                    { 3, 1 } };
 
    // Sort the major Diagonal
    sortDiagonal(a, 2, 2);
 
    return 0;
}

Java

// Java implementation to sort the
// major diagonal of the matrix
class GFG{
  
// Function to sort the major
// diagonal of the matrix
static void sortDiagonal(int a[][], int M, int N)
{
    // Loop to find the ith minimum
    // element from the major diagonal
    for (int i = 0; i < M; i++) {
        int sm = a[i][i];
        int pos = i;
          
        // Loop to find the minimum
        // element from the unsorted matrix
        for (int j = i + 1; j < N; j++) {
            if (sm > a[j][j]) {
                sm = a[j][j];
                pos = j;
            }
        }
          
        // Swap to put the minimum
        // element at the beginning of
        // the major diagonal of matrix
        swap(a, i, i, pos, pos);
    }
  
    // Loop to print the matrix
    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++)
            System.out.print(a[i][j]+ " ");
        System.out.println();
    }
}
  
static void swap(int[][] a, int i, int i2, int pos, int pos2) {
    int temp = a[i][i2];
    a[i][i2] = a[pos][pos2];
    a[pos][pos2] = temp;   
}
 
// Driven Code
public static void main(String[] args)
{
    int a[][] = { { 4, 2 },
                    { 3, 1 } };
  
    // Sort the major Diagonal
    sortDiagonal(a, 2, 2);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to sort the
# major diagonal of the matrix
 
# Function to sort the major
# diagonal of the matrix
def sortDiagonal(a, M, N):
     
    # Loop to find the ith minimum
    # element from the major diagonal
    for i in range(M):
        sm = a[i][i]
        pos = i
         
        # Loop to find the minimum
        # element from the unsorted matrix
        for j in range(i + 1 , N):
            if (sm > a[j][j]):
                sm = a[j][j]
                pos = j
                 
        # Swap to put the minimum
        # element at the beginning of
        # the major diagonal of matrix
        a[i][i], a[pos][pos] = a[pos][pos] , a[i][i]
         
    # Loop to print the matrix
    for i in range(M):
        for j in range(N):
            print(a[i][j],end=" ")
        print()
 
# Driven Code
a = [[4, 2],[3, 1]]
 
# Sort the major Diagonal
sortDiagonal(a, 2, 2)
 
# This code is contributed by shubhamsingh10

C#

// C# implementation to sort the
// major diagonal of the matrix
using System;
 
class GFG{
   
// Function to sort the major
// diagonal of the matrix
static void sortDiagonal(int[,]a, int M, int N)
{
    // Loop to find the ith minimum
    // element from the major diagonal
    for (int i = 0; i < M; i++) {
        int sm = a[i, i];
        int pos = i;
           
        // Loop to find the minimum
        // element from the unsorted matrix
        for (int j = i + 1; j < N; j++) {
            if (sm > a[j, j]) {
                sm = a[j, j];
                pos = j;
            }
        }
           
        // Swap to put the minimum
        // element at the beginning of
        // the major diagonal of matrix
        swap(a, i, i, pos, pos);
    }
   
    // Loop to print the matrix
    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++)
            Console.Write(a[i, j]+ " ");
        Console.WriteLine();
    }
}
   
static void swap(int[,] a, int i, int i2, int pos, int pos2) {
    int temp = a[i, i2];
    a[i, i2] = a[pos, pos2];
    a[pos, pos2] = temp;   
}
  
// Driven Code
public static void Main(String[] args)
{
    int[,]a = { { 4, 2 },
                    { 3, 1 } };
   
    // Sort the major Diagonal
    sortDiagonal(a, 2, 2);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation to sort the
// major diagonal of the matrix
 
// Function to sort the major
// diagonal of the matrix
function sortDiagonal(a, M, N)
{
    // Loop to find the ith minimum
    // element from the major diagonal
    for (var i = 0; i < M; i++) {
        var sm = a[i][i];
        var pos = i;
         
        // Loop to find the minimum
        // element from the unsorted matrix
        for (var j = i + 1; j < N; j++) {
            if (sm > a[j][j]) {
                sm = a[j][j];
                pos = j;
            }
        }
         
        // Swap to put the minimum
        // element at the beginning of
        // the major diagonal of matrix
        temp = a[i][i]
        a[i][i] = a[pos][pos]
        a[pos][pos] = temp
    }
 
    // Loop to print the matrix
    for (var i = 0; i < M; i++) {
        for (var j = 0; j < N; j++)
            document.write( a[i][j] + " ");
        document.write("<br>");
    }
}
 
// Driven Code
var a = [ [ 4, 2 ],
                [ 3, 1 ] ];
// Sort the major Diagonal
sortDiagonal(a, 2, 2);
 
// This code is contributed by noob2000.

输出：

1 2 
3 4

性能分析：

• 时间复杂度： O(min(M, N) ² )
• 辅助空间： O(1)

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