将一个二维数组（方阵）顺时针旋转90°_交换思想

忘是亡心i 2022-05-25 07:15 236阅读 0赞

#### **做题思路：** ####

![这里写图片描述][SouthEast]

#### **程序清单：** ####

#include <iostream>
    #include <iomanip>
    using namespace std;
    
    #define N 4
    //将一个二维N*N矩阵顺时针旋转90°
    //先转置，再逆序
    /* 1 2 3 4 1 5 9 13 13 9 5 1 5 6 7 8 转置 2 6 10 14 逆序 14 10 6 2 9 10 11 12 =======> 3 7 11 15 ========> 15 11 7 3 13 14 15 16 4 8 12 16 16 12 8 4 */
    void show(int a[][N])
    {
        for (int i = 0; i < N; ++i)
        {
            for (int j = 0; j < N; ++j)
            {
                cout << setw(4) << a[i][j] << " ";
            }
            cout << endl;
        }
        cout << endl;
    }
    void rotate(int a[][N])
    {
        //转置
        for (int i = 0; i < N; ++i)
        {
            for (int j = 0; j < i; ++j)
            {
                int temp = 0;
                //沿主对角线对称位置上的元素进行交换
                temp = a[i][j];
                a[i][j] = a[j][i];
                a[j][i] = temp;
            }
        }
        cout << "转置后结果如下：" << endl;
        show(a);
        //逆序：对一维数组而言，将每一行第i个元素与第N-i-1个元素进行交换
        for (int i = 0; i < N; ++i)
        {
            for (int j = N - 1; j >= N / 2; --j) //注意j循环的条件为 j>=N/2
            {
                int temp = 0;
                temp = a[i][j];   //第i行末元素
                a[i][j] = a[i][N - j - 1];
                a[i][N - j - 1] = temp;
            }
        }
        cout << "逆序后得到旋转后的最终结果如下：" << endl;
        show(a);
    }
    int main(void)
    {
        int a[N][N] = {
            {
        1,2,3,4},
            {
        5,6,7,8},
            {
        9,10,11,12},
            {
        13,14,15,16} };
        cout << "原矩阵如下所示：" << endl;
        show(a);
        rotate(a);
        return 0;
    }

#### **运行结果：** ####

![这里写图片描述][SouthEast 1]

[SouthEast]: /images/20220525/9e6e6f62f0494abda5f38517fc7f2ad1.png
[SouthEast 1]: /images/20220525/f177748d8b6b4666ba8e4fd9b5fd698e.png