检查矩阵的所有行是否相互循环旋转

给定一个 n*n 大小的矩阵，任务是找出所有行是否是彼此的圆形旋转。

例子：

Input: mat[][] = 1, 2, 3
                 3, 1, 2
                 2, 3, 1
Output:  Yes
All rows are rotated permutation
of each other.

Input: mat[3][3] = 1, 2, 3
                   3, 2, 1
                   1, 3, 2
Output:  No
Explanation : As 3, 2, 1 is not a rotated or 
circular permutation of 1, 2, 3

这个想法是基于下面的文章。
检查字符串是否相互旋转的程序

脚步 ：

1. 创建一个由第一行元素组成的字符串并将该字符串与其自身连接，以便可以有效地执行字符串搜索操作。让这个字符串为 str_cat。
2. 遍历所有剩余的行。对于要遍历的每一行，创建一个包含当前行元素的字符串str_curr。如果 str_curr 不是 str_cat 的子字符串，则返回 false。
3. 返回真。

下面是上述步骤的实现。

C++

// C++ program to check if all rows of a matrix
// are rotations of each other
#include 
using namespace std;
const int MAX = 1000;
 
// Returns true if all rows of mat[0..n-1][0..n-1]
// are rotations of each other.
bool isPermutedMatrix( int mat[MAX][MAX], int n)
{
    // Creating a string that contains elements of first
    // row.
    string str_cat = "";
    for (int i = 0 ; i < n ; i++)
        str_cat = str_cat + "-" + to_string(mat[0][i]);
 
    // Concatenating the string with itself so that
    // substring search operations can be performed on
    // this
    str_cat = str_cat + str_cat;
 
    // Start traversing remaining rows
    for (int i=1; i

---

Java

// Java program to check if all rows of a matrix
// are rotations of each other
class GFG
{
 
    static int MAX = 1000;
 
    // Returns true if all rows of mat[0..n-1][0..n-1]
    // are rotations of each other.
    static boolean isPermutedMatrix(int mat[][], int n)
    {
        // Creating a string that contains
        // elements of first row.
        String str_cat = "";
        for (int i = 0; i < n; i++)
        {
            str_cat = str_cat + "-" + String.valueOf(mat[0][i]);
        }
 
        // Concatenating the string with itself
        // so that substring search operations 
        // can be performed on this
        str_cat = str_cat + str_cat;
 
        // Start traversing remaining rows
        for (int i = 1; i < n; i++)
        {
            // Store the matrix into vector in the form
            // of strings
            String curr_str = "";
            for (int j = 0; j < n; j++)
            {
                curr_str = curr_str + "-" + String.valueOf(mat[i][j]);
            }
 
            // Check if the current string is present in
            // the concatenated string or not
            if (str_cat.contentEquals(curr_str))
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Drivers code
    public static void main(String[] args)
    {
        int n = 4;
        int mat[][] = {{1, 2, 3, 4},
        {4, 1, 2, 3},
        {3, 4, 1, 2},
        {2, 3, 4, 1}
        };
        if (isPermutedMatrix(mat, n))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to check if all rows
# of a matrix are rotations of each other
 
MAX = 1000
 
# Returns true if all rows of mat[0..n-1][0..n-1]
# are rotations of each other.
def isPermutedMatrix(mat, n) :
     
    # Creating a string that contains
    # elements of first row.
    str_cat = ""
    for i in range(n) :
        str_cat = str_cat + "-" + str(mat[0][i])
 
    # Concatenating the string with itself
    # so that substring search operations
    # can be performed on this
    str_cat = str_cat + str_cat
 
    # Start traversing remaining rows
    for i in range(1, n) :
         
        # Store the matrix into vector
        # in the form of strings
        curr_str = ""
         
        for j in range(n) :
            curr_str = curr_str + "-" + str(mat[i][j])
 
        # Check if the current string is present
        # in the concatenated string or not
        if (str_cat.find(curr_str)) :
            return True
             
    return False
 
# Driver code
if __name__ == "__main__" :
    n = 4
    mat = [[1, 2, 3, 4],
           [4, 1, 2, 3],
           [3, 4, 1, 2],
           [2, 3, 4, 1]]
     
    if (isPermutedMatrix(mat, n)):
        print("Yes")
    else :
        print("No")
         
# This code is contributed by Ryuga

C#

// C# program to check if all rows of a matrix
// are rotations of each other
using System;
 
class GFG
{
 
    //static int MAX = 1000;
 
    // Returns true if all rows of mat[0..n-1,0..n-1]
    // are rotations of each other.
    static bool isPermutedMatrix(int [,]mat, int n)
    {
        // Creating a string that contains
        // elements of first row.
        string str_cat = "";
        for (int i = 0; i < n; i++)
        {
            str_cat = str_cat + "-" + mat[0,i].ToString();
        }
 
        // Concatenating the string with itself
        // so that substring search operations
        // can be performed on this
        str_cat = str_cat + str_cat;
 
        // Start traversing remaining rows
        for (int i = 1; i < n; i++)
        {
            // Store the matrix into vector in the form
            // of strings
            string curr_str = "";
            for (int j = 0; j < n; j++)
            {
                curr_str = curr_str + "-" + mat[i,j].ToString();
            }
 
            // Check if the current string is present in
            // the concatenated string or not
            if (str_cat.Equals(curr_str))
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    static void Main()
    {
        int n = 4;
        int [,]mat = {{1, 2, 3, 4},
        {4, 1, 2, 3},
        {3, 4, 1, 2},
        {2, 3, 4, 1}
        };
         
        if (isPermutedMatrix(mat, n))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
 
/* This code contributed by mits */

Javascript

 
// Javascript program to check if all rows of a matrix
// are rotations of each other
     
// Returns true if all rows of mat[0..n-1][0..n-1]
// are rotations of each other.
function isPermutedMatrix(mat, n)
{
     
    // Creating a string that contains
    // elements of first row.
    let str_cat = "";
    for (let i = 0; i < n; i++)
    {
        str_cat = str_cat + "-" +
                  (mat[0][i]).toString();
    }
     
    // Concatenating the string with itself
    // so that substring search operations 
    // can be performed on this
    str_cat = str_cat + str_cat;
 
    // Start traversing remaining rows
    for(let i = 1; i < n; i++)
    {
         
        // Store the matrix into vector in the form
        // of strings
        let curr_str = "";
        for(let j = 0; j < n; j++)
        {
            curr_str = curr_str + "-" +
                       (mat[i][j]).toString();
        }
         
        // Check if the current string is present in
        // the concatenated string or not
        if (str_cat.includes(curr_str))
        {
            return true;
        }
    }
    return false;
}
 
// Drivers code
let n = 4;
let mat = [ [ 1, 2, 3, 4 ],
            [ 4, 1, 2, 3 ],
            [ 3, 4, 1, 2 ],
            [ 2, 3, 4, 1 ] ];
             
if (isPermutedMatrix(mat, n))
    document.write("Yes")
else
    document.write("No")
 
// This code is contributed by rag2127
 

输出：

Yes