旋转的矩阵 顺时针逆时针打印C++

#include 
#include
using namespace std;
int main()
{int cal = 1;//记录第几圈 奇数就是顺指针 偶数就是逆时针vector>V;//V是二维数组int row, col;cin >> row >> col;for (int i = 0; i < row; i++){vectorv;for (int j = 0; j < col; j++){int val;cin >> val;v.push_back(val);}V.push_back(v);}//输出整个矩阵//for (int i = 0; i < row; i++)//{//	for (int j = 0; j < col; j++)//	{//		cout << V[i][j]<<" ";//	}//	cout << endl;////}int top = 0, bottle = row - 1, left = 0, right = col - 1;//记录矩阵边界 遍历一行/一列 边界减小//这个循环条件为什么要+1 自己模拟一下 如果不+1 一次遍历完之后 会有一行还没遍历上就退出循环了while (top != bottle + 1 || right != left - 1){//顺时针if (cal % 2 != 0){for (int i = left; i <= right; i++)//从左到右{cout << V[top][i] << " ";}top++;if (top == bottle + 1 || right == left - 1)break;for (int i = top; i <= bottle; i++)//从上到下{cout << V[i][right] << " ";}right--;if (top == bottle + 1 || right == left - 1)break;for (int i = right; i >= left; i--)//从右到左{cout << V[bottle][i] << " ";}bottle--;if (top == bottle + 1 || right == left - 1)break;for (int i = bottle; i >= top; i--)//从下到上{cout << V[i][left] << " ";}left++;cal++;}//逆时针 (偶数圈 如果还可以继续遍历的话）else{for (int i = top; i <= bottle; i++)//从上到下{cout << V[i][left] << " ";}left++;if (top == bottle + 1 || right == left - 1)break;for (int i = left; i <= right; i++)//从左到右{cout << V[bottle][i] << " ";}bottle--;if (top == bottle + 1 || right == left - 1)break;for (int i = bottle; i >= top; i--)//从下到上{cout << V[i][right] << " ";}right--;if (top == bottle + 1 || right == left - 1)break;for (int i = right; i >= left; i--)//从右到左{cout << V[top][i] << " ";}top++;cal++;}}return 0;
}

 代码写的巨冗余....但AC了就懒得再改进了（笑）

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