检查矩阵是否对称的程序

如果矩阵的转置与给定矩阵相同，则称方阵为对称矩阵。可以通过将行改为列、将列改为行来获得对称矩阵。

例子：

Input : 1 2 3
        2 1 4
        3 4 3
Output : Yes

Input : 3 5 8
        3 4 7
        8 5 3
Output : No

一个简单的解决方案是执行以下操作。

1）创建给定矩阵的转置。
2）检查转置矩阵和给定矩阵是否相同，

C++

// Simple c++ code for check a matrix is
// symmetric or not.
#include 
using namespace std;
 
const int MAX = 100;
 
// Fills transpose of mat[N][N] in tr[N][N]
void transpose(int mat[][MAX], int tr[][MAX], int N)
{
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            tr[i][j] = mat[j][i];
}
 
// Returns true if mat[N][N] is symmetric, else false
bool isSymmetric(int mat[][MAX], int N)
{
    int tr[N][MAX];
    transpose(mat, tr, N);
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            if (mat[i][j] != tr[i][j])
                return false;
    return true;
}
 
// Driver code
int main()
{
    int mat[][MAX] = { { 1, 3, 5 },
                       { 3, 2, 4 },
                       { 5, 4, 1 } };
 
    if (isSymmetric(mat, 3))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Simple java code for check a matrix is
// symmetric or not.
 
import java.io.*;
 
class GFG {
     
 
 
 static int  MAX = 100;
 
// Fills transpose of mat[N][N] in tr[N][N]
 static void transpose(int mat[][], int tr[][], int N)
{
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            tr[i][j] = mat[j][i];
}
 
// Returns true if mat[N][N] is symmetric, else false
 static boolean isSymmetric(int mat[][], int N)
{
    int tr[][] = new int[N][MAX];
    transpose(mat, tr, N);
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            if (mat[i][j] != tr[i][j])
                return false;
    return true;
}
 
// Driver code
    public static void main (String[] args)
 {
         
        int mat[][] = { { 1, 3, 5 },
                    { 3, 2, 4 },
                    { 5, 4, 1 } };
 
    if (isSymmetric(mat, 3))
        System.out.println( "Yes");
    else
        System.out.println ( "No");
     
    }
}

Python

# Simple Python code for check a matrix is
# symmetric or not.
  
# Fills transpose of mat[N][N] in tr[N][N]
def transpose(mat, tr, N):
    for i in range(N):
        for j in range(N):
            tr[i][j] = mat[j][i]
  
# Returns true if mat[N][N] is symmetric, else false
def isSymmetric(mat, N):
     
    tr = [ [0 for j in range(len(mat[0])) ] for i in range(len(mat)) ]
    transpose(mat, tr, N)
    for i in range(N):
        for j in range(N):
            if (mat[i][j] != tr[i][j]):
                return False
    return True
  
# Driver code
mat = [ [ 1, 3, 5 ], [ 3, 2, 4 ], [ 5, 4, 1 ] ]
if (isSymmetric(mat, 3)):
    print "Yes"
else:
    print "No"
 
# This code is contributed by Sachin Bisht

C#

// Simple C# code for check a matrix is
// symmetric or not.
 
using System;
 
class GFG {
     
    static int MAX = 100;
     
    // Fills transpose of mat[N][N] in tr[N][N]
    static void transpose(int [,]mat, int [,]tr, int N)
    {
        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                tr[i,j] = mat[j,i];
    }
     
    // Returns true if mat[N][N] is symmetric, else false
    static bool isSymmetric(int [,]mat, int N)
    {
        int [,]tr = new int[N,MAX];
        transpose(mat, tr, N);
        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                if (mat[i,j] != tr[i,j])
                    return false;
        return true;
    }
     
    // Driver code
    public static void Main ()
    {
             
        int [,]mat = { { 1, 3, 5 },
                        { 3, 2, 4 },
                        { 5, 4, 1 } };
     
        if (isSymmetric(mat, 3))
            Console.WriteLine( "Yes");
        else
            Console.WriteLine( "No");
         
      }
}
 
// This code is contributed by vt_m.

Javascript

 
// Simple javascript code for check
// a matrix is symmetric or not.
let  MAX = 100;
 
// Fills transpose of mat[N][N] in tr[N][N]
function transpose(mat, tr, N)
{
    for(let i = 0; i < N; i++)
        for(let j = 0; j < N; j++)
            tr[i][j] = mat[j][i];
}
 
// Returns true if mat[N][N] is
// symmetric, else false
function isSymmetric(mat, N)
{
    let tr = new Array(N);
    for(let i = 0; i < N; i++)
    {
        tr[i] = new Array(MAX);
    }
    transpose(mat, tr, N);
     
    for(let i = 0; i < N; i++)
        for(let j = 0; j < N; j++)
            if (mat[i][j] != tr[i][j])
                return false;
                 
    return true;
}
 
// Driver code
let mat = [ [ 1, 3, 5 ],
            [ 3, 2, 4 ],
            [ 5, 4, 1 ] ];
 
if (isSymmetric(mat, 3))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by decode2207
 

输出 ：

Yes

时间复杂度：O(N x N)
辅助空间：O(N x N)

检查矩阵是否对称的有效解决方案是在不创建转置的情况下比较矩阵元素。我们基本上需要比较 mat[i][j] 和 mat[j][i]。

C++

// Efficient c++ code for check a matrix is
// symmetric or not.
#include 
using namespace std;
 
const int MAX = 100;
 
// Returns true if mat[N][N] is symmetric, else false
bool isSymmetric(int mat[][MAX], int N)
{
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            if (mat[i][j] != mat[j][i])
                return false;
    return true;
}
 
// Driver code
int main()
{
    int mat[][MAX] = { { 1, 3, 5 },
                       { 3, 2, 4 },
                       { 5, 4, 1 } };
 
    if (isSymmetric(mat, 3))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Efficient Java code for check a matrix is
// symmetric or no
 
import java.io.*;
 
class GFG {
    
 
static int MAX = 100;
 
// Returns true if mat[N][N]
// is symmetric, else false
 static boolean isSymmetric(int mat[][], int N)
{
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            if (mat[i][j] != mat[j][i])
                return false;
    return true;
}
 
// Driver code
     
    public static void main (String[] args)
 {
            int mat[][] = { { 1, 3, 5 },
                    { 3, 2, 4 },
                    { 5, 4, 1 } };
 
    if (isSymmetric(mat, 3))
        System.out.println(  "Yes");
    else
         
        System.out.println("NO");
         
    }
}
// This article is contributed by vt_m.

Python

# Efficient Python code for check a matrix is
# symmetric or not.
 
# Returns true if mat[N][N] is symmetric, else false
def isSymmetric(mat, N):
    for i in range(N):
        for j in range(N):
            if (mat[i][j] != mat[j][i]):
                return False
    return True
  
# Driver code
mat = [ [ 1, 3, 5 ], [ 3, 2, 4 ], [ 5, 4, 1 ] ]
if (isSymmetric(mat, 3)):
    print "Yes"
else:
    print "No"
 
# This code is contributed by Sachin Bisht

C＃

// Efficient C# code for check a matrix is
// symmetric or no
 
using System;
 
class GFG
{
    //static int MAX = 100;
     
    // Returns true if mat[N][N]
    // is symmetric, else false
    static bool isSymmetric(int [,]mat, int N)
    {
        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                if (mat[i, j] != mat[j, i])
                    return false;
        return true;
    }
     
    // Driver code
    public static void Main ()
    {
        int [,]mat = { { 1, 3, 5 },
                    { 3, 2, 4 },
                    { 5, 4, 1 } };
 
        if (isSymmetric(mat, 3))
            Console.WriteLine( "Yes");
        else
             
            Console.WriteLine("NO");
         
    }
}
 
// This code is contributed by vt_m.

Javascript

    // Efficient Javascript code for check a matrix is symmetric or no
     
    let MAX = 100;
   
    // Returns true if mat[N][N]
    // is symmetric, else false
    function isSymmetric(mat, N)
    {
        for (let i = 0; i < N; i++)
            for (let j = 0; j < N; j++)
                if (mat[i][j] != mat[j][i])
                    return false;
        return true;
    }
     
    let mat = [ [ 1, 3, 5 ],
               [ 3, 2, 4 ],
               [ 5, 4, 1 ] ];
   
    if (isSymmetric(mat, 3))
        document.write(  "Yes");
    else
           
        document.write("NO");
 

输出：

Yes

时间复杂度：O(N x N)
辅助空间：O(1)