顺时针打印nxn的矩阵

对于如下的矩阵进行顺时针的打印：

1 2 3

4 5 6

7 8 9

打印结果是：1,2,3,6,9,8,7,4,5这个打印很简单，就是按圈进行打印，看这个矩阵的维度是3，那么一共有一圈，和一个独立的点5，如果维度是4，那么就有两圈，没有独立的点，独立点进行单独处理，那么按圈进行打印，就要简单的多。自己拿出草稿纸画画就好，关键点是第几圈要有记录。

#include<iostream>using std::cin;using std::cout;using std::endl;const int N = 5;void clockwisePrint(int** arr, int dimension) {int level, col, row;for (level = 1; level <= dimension / 2; level++) {for (col = level - 1; col < dimension - level; col++)cout << arr[level - 1][col] << " ";for (row = level - 1; row < dimension - level; row++)cout << arr[row][dimension -level] << " ";for (col = dimension - level; col > level - 1; col--)cout << arr[dimension - level][col] << " ";for (row = dimension - level; row > level - 1; row--)cout << arr[row][level - 1] << " ";}if (1 == dimension % 2)cout << arr[dimension / 2][dimension / 2];cout << endl;}int main(int argc, char* argv[]) {int** arr = (int**)malloc(sizeof(int*) * N);int i, j;for (i = 0; i < N; i++)arr[i] = (int*)malloc(sizeof(int) * N);int count = 1;for (i = 0; i < N; i++) {for (j = 0; j < N; j++) {arr[i][j] = count;count++;}}clockwisePrint(arr, N);for (i = 0; i < N; i++)free(arr[i]);free(arr);cin.get();return 0;}