检查行反转后矩阵是否保持不变

给定一个 NxN 矩阵。任务是检查在反转给定矩阵的所有行后，矩阵是否保持不变。

例子：

Input : N = 3
1 2 1
2 2 2
3 4 3 
Output : Yes 
If all the rows are reversed then matrix will become:
1 2 1
2 2 2
3 4 3 
which is same.

Input : N = 3
1 2 2
2 2 2
3 4 3
Output : No

方法：

一个最重要的观察是矩阵在行反转后是相同的，每一行都必须是回文的。
现在检查一行是否是回文，维护两个指针，一个指向开始，另一个指向行尾。开始比较存在的值并执行 start++ 和 end–。重复这个过程，直到所有元素都被检查到行的中间。如果每个步骤的元素都相同，则该行是回文，否则不是。
如果任何一行不是回文，那么答案是否定的。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
#define ll long long int
using namespace std;
 
// Function to check Palindromic Condition
void specialMatrix(int matrix[3][3], int N)
{
    for (int i = 0; i < N; i++) {
 
        // Pointer to start of row
        int start = 0;
 
        // Pointer to end of row
        int end = N - 1;
 
        while (start <= end) {
 
            // Single Mismatch means row is not palindromic
            if (matrix[i][start] != matrix[i][end]) {
                cout << "No" << endl;
                return;
            }
 
            start++;
            end--;
        }
    }
 
    cout << "Yes" << endl;
    return;
}
 
// Driver Code
int main()
{
    int matrix[3][3] = { { 1, 2, 1 },
                         { 2, 2, 2 },
                         { 3, 4, 3 } };
    int N = 3;
    specialMatrix(matrix, N);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
 
// Function to check Palindromic Condition
static void specialMatrix(int matrix[][], int N)
{
    for (int i = 0; i < N; i++)
    {
 
        // Pointer to start of row
        int start = 0;
 
        // Pointer to end of row
        int end = N - 1;
 
        while (start <= end)
        {
 
            // Single Mismatch means
            // row is not palindromic
            if (matrix[i][start] != matrix[i][end])
            {
                System.out.println("No");
                return;
            }
 
            start++;
            end--;
        }
    }
 
    System.out.println("Yes");
    return;
}
 
// Driver Code
public static void main(String[] args)
{
    int matrix[][] = { { 1, 2, 1 },
                        { 2, 2, 2 },
                        { 3, 4, 3 } };
    int N = 3;
    specialMatrix(matrix, N);
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python 3 implementation of the above approach
 
# Function to check Palindromic Condition
def specialMatrix(matrix, N):
    for i in range(N):
         
        # Pointer to start of row
        start = 0
 
        # Pointer to end of row
        end = N - 1
 
        while (start <= end):
             
            # Single Mismatch means row is not palindromic
            if (matrix[i][start] != matrix[i][end]):
                print("No")
                return
 
            start += 1
            end -= 1
 
    print("Yes")
    return
 
# Driver Code
if __name__ == '__main__':
    matrix = [[1, 2, 1], [2, 2, 2], [3, 4, 3]]
    N = 3
    specialMatrix(matrix, N)
     
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
 
class GFG
{
 
// Function to check Palindromic Condition
static void specialMatrix(int [,]matrix, int N)
{
    for (int i = 0; i < N; i++)
    {
 
        // Pointer to start of row
        int start = 0;
 
        // Pointer to end of row
        int end = N - 1;
 
        while (start <= end)
        {
 
            // Single Mismatch means
            // row is not palindromic
            if (matrix[i, start] != matrix[i, end])
            {
                Console.WriteLine("No");
                return;
            }
 
            start++;
            end--;
        }
    }
    Console.WriteLine("Yes");
    return;
}
 
// Driver Code
public static void Main(String[] args)
{
    int [,]matrix = { { 1, 2, 1 },
                        { 2, 2, 2 },
                        { 3, 4, 3 } };
    int N = 3;
    specialMatrix(matrix, N);
}
}
 
// This code has been contributed by 29AjayKumar

PHP

Javascript

输出：

Yes