# 检查矩阵的所有行是否都是彼此旋转 > 原文： [https://www.geeksforgeeks.org/check-rows-matrix-circular-rotations/](https://www.geeksforgeeks.org/check-rows-matrix-circular-rotations/) 给定 n * n 大小的矩阵，任务是查找所有行是否都是彼此的圆周旋转。 **示例**： ``` 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。 [用于检查字符串是否彼此旋转的程序](https://www.geeksforgeeks.org/a-program-to-check-if-strings-are-rotations-of-each-other/) **步骤**： 1. 创建一个由第一行元素组成的字符串，并将其与自身连接在一起，以便可以高效地执行字符串搜索操作。 将此字符串设为 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 true。 下面是上述步骤的实现。 ## C++ ```cpp // C++ program to check if all rows of a matrix // are rotations of each other #include <bits/stdc++.h> 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<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 + "-" + to_string(mat[i][j]);         // Check if the current string is present in         // the concatenated string or not         if (str_cat.find(curr_str) == string::npos)             return false;     }     return true; } // Drivers code int main() {     int n = 4 ;     int mat[MAX][MAX] = {{1, 2, 3, 4},         {4, 1, 2, 3},         {3, 4, 1, 2},         {2, 3, 4, 1}     };     isPermutedMatrix(mat, n)? cout << "Yes" :                               cout << "No";     return 0; } ```