剑指 Offer 29. 顺时针打印矩阵

思路：
　　　用模拟的方式，非常妙，学习下方向存储矩阵

int directions[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};　//每一行代表一个方向，然后是x和y控制上下和左右；这边是右下，左上

class Solution {
private:
    int directions[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if (matrix.size() == 0) {
            return {};
        }
        int rows = matrix.size(), columns = matrix[0].size();
        vector<vector<bool>> visited(rows, vector<bool>(columns));
        int total = rows * columns;
        vector<int> order(total);

        int row = 0, column = 0;
        int directionIndex = 0;
        for (int i = 0; i < total; i++) {
            order[i] = matrix[row][column];  //行和列在上一次已经更新了，这次只是输出
            visited[row][column] = true;  //确认访问。
            int nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1];
            if (nextRow < 0 || nextRow >= rows || nextColumn < 0 || nextColumn >= columns || visited[nextRow][nextColumn]) {
                directionIndex = (directionIndex + 1) % 4; //妙啊啊啊啊 通过visit缩小空间
            }
            //上面验证了next的步伐是没问题的，有问题就改变了
            row += directions[directionIndex][0];    //然后这边就更新了
            column += directions[directionIndex][1];
        }
        return order;
    }
};