leetcode-顺时针打印矩阵

法一：辅助矩阵与方向数组

class Solution {
private:
	static constexpr int directions[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
    	if(matrix.size() == 0 || matrix[0].size() == 0) return {};
		
		int rows = matrix.size();
		int cols = matrix[0].size();
		// 创建一个辅助的矩阵 
		vector<vector<bool>> visited(rows, vector<bool>(cols));
		int total = rows * cols; 
		vector<int> order(total);
		
		int row = 0, col = 0;
		int directionIndex = 0;
		for (int i = 0; i < total; i++){
			order[i] = matrix[row][col];
			visited[row][col] = true;
			int nextRow = row + directions[directionIndex][0], nextColumn = col + directions[directionIndex][1];
			if (nextRow < 0 || nextRow >= rows ||  nextColumn < 0 || nextColumn >= cols || visited[nextRow][nextColumn]){
				directionIndex = (directionIndex + 1) % 4;
			}
			row += directions[directionIndex][0];
			col += directions[directionIndex][1];	
		}
    	
		return order;
	}
};

法二：定位遍历

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if (matrix.size() == 0 || matrix[0].size() == 0) {
            return {};
        }

        int rows = matrix.size(), columns = matrix[0].size();
        vector<int> order;
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                order.push_back(matrix[top][column]);
            }
            for (int row = top + 1; row <= bottom; row++) {
                order.push_back(matrix[row][right]);
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    order.push_back(matrix[bottom][column]);
                }
                for (int row = bottom; row > top; row--) {
                    order.push_back(matrix[row][left]);
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return order;
    }
};

总结：

1. 方向数组的构建和使用
   

int xy[8][2]={{0,1},{0,-1},{1,0},{-1,0},{1,-1},{1,1},{-1,-1},{-1,1}};
for(int i=0;i<8;i++){
        int nx=x+xy[i][0];
        int ny=y+xy[i][1];
}

2. 创建一个二维矩阵的方法（必须要初始化，否则会报错）

vector<vector<bool>> visited(rows, vector<bool>(cols));

// 二维数组
int numRows=10,zone=5;//层数，每层需要的空间
vector<vector<int>> vec(numRows, vector<int>());//初始层数，赋值
for (int i = 0; i < numRows; i++) {
	vec[i].resize(zone);
}