54. Spiral Matrix 螺旋矩阵

将一个二维矩阵，螺旋方式遍历。

Example 1:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

思考：本题的难点在于以螺旋的方式访问。需要考虑各个方向和是否访问过，何时旋转方向继续向前。

解法：

方法1：使用辅助空间记录是否已经访问过。

    vector spiralOrder(vector>& matrix) {
        vector res;
        if (matrix.empty()) { return res;}
        
        int rows = matrix.size();
        int cols = matrix[0].size();
        
        vector> visited(rows, vector(cols, false));
        int len = rows * cols;
        int r = 0, c = 0;
        int j = 0;
        
        int dr[4] = {0, 1, 0, -1};
        int dc[4] = {1, 0, -1, 0};
        
        
        for (int i = 0; i < len; ++i) {
            res.push_back(matrix[r][c]);
            visited[r][c] = true;
            
            int n_r = r + dr[j];
            int n_c = c + dc[j];
            if ((n_r >= 0 && n_r < rows) && (n_c >= 0 && n_c < cols) && (!visited[n_r][n_c])) {
                r = n_r;
                c = n_c;
            } else {
                j = (j + 1) % 4;
                r += dr[j];
                c += dc[j];
            }
        }
        return res;
    }

方法2：直接遍历。需要注意while，for循环的比较运算符。

    vector spiralOrder(vector>& matrix) {
        vector res;
        if (matrix.empty() || matrix[0].empty()) { return res; }
        
        int r1 = 0;
        int r2 = matrix.size() - 1;
        
        int c1 = 0;
        int c2 = matrix[0].size() - 1;
        
        while(r1 <= r2 && c1 <= c2) {
            for (int c = c1; c <= c2; ++c) { res.push_back(matrix[r1][c]); }
            for (int r = r1 + 1; r <= r2; ++r) { res.push_back(matrix[r][c2]); }

            if (r1 < r2 && c1 < c2) {
                for (int c = c2 - 1; c > c1; --c) { res.push_back(matrix[r2][c]); }
                for (int r = r2; r > r1; --r) { res.push_back(matrix[r][c1]); }
            }
            
            r1++;
            r2--;
            c1++;
            c2--;
        }
        return res;
    }