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）