# 旋转矩阵元素 > 原文： [https://www.geeksforgeeks.org/rotate-matrix-elements/](https://www.geeksforgeeks.org/rotate-matrix-elements/) 给定一个矩阵，顺时针旋转其中的元素。 **示例**： ``` Input 1 2 3 4 5 6 7 8 9 Output: 4 1 2 7 5 3 8 9 6 For 4*4 matrix Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12 ``` 想法是使用类似于[程序的循环以螺旋形式](https://www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/)打印矩阵。 从最外层开始，一圈一圈地旋转所有元素环。 要旋转戒指，我们需要执行以下操作。 1）移动第一行的元素。 2）移动最后一列的元素。 3）移动底部一行的元素。 4）移动第一列的元素。 有内圈时，对内圈重复上述步骤。 以下是上述想法的实现。 感谢 Gaurav Ahirwar 提供以下解决方案的建议。 ## C/C++ ``` // C++ program to rotate a matrix #include <bits/stdc++.h> #define R 4 #define C 4 using namespace std; // A function to rotate a matrix mat[][] of size R x C. // Initially, m = R and n = C void rotatematrix(int m, int n, int mat[R][C]) {     int row = 0, col = 0;     int prev, curr;     /*        row - Staring row index        m - ending row index        col - starting column index        n - ending column index        i - iterator     */     while (row < m && col < n)     {         if (row + 1 == m || col + 1 == n)             break;         // Store the first element of next row, this         // element will replace first element of current         // row         prev = mat[row + 1][col];          /* Move elements of first row from the remaining rows */         for (int i = col; i < n; i++)         {             curr = mat[row][i];             mat[row][i] = prev;             prev = curr;         }         row++;         /* Move elements of last column from the remaining columns */         for (int i = row; i < m; i++)         {             curr = mat[i][n-1];             mat[i][n-1] = prev;             prev = curr;         }         n--;          /* Move elements of last row from the remaining rows */         if (row < m)         {             for (int i = n-1; i >= col; i--)             {                 curr = mat[m-1][i];                 mat[m-1][i] = prev;                 prev = curr;             }         }         m--;         /* Move elements of first column from the remaining rows */         if (col < n)         {             for (int i = m-1; i >= row; i--)             {                 curr = mat[i][col];                 mat[i][col] = prev;                 prev = curr;             }         }         col++;     }     // Print rotated matrix     for (int i=0; i<R; i++)     {         for (int j=0; j<C; j++)           cout << mat[i][j] << " ";         cout << endl;     } } /* Driver program to test above functions */ int main() {     // Test Case 1     int a[R][C] = { {1,  2,  3,  4},         {5,  6,  7,  8},         {9,  10, 11, 12},         {13, 14, 15, 16}  };     // Tese Case 2     /* int a[R][C] = {{1, 2, 3},                       {4, 5, 6},                       {7, 8, 9}                      };      */  rotatematrix(R, C, a);     return 0; } ```