将一个二维数组（方阵）顺时针旋转90°_交换思想

做题思路：

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：