给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

示例 1：

**输入：**matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]

示例 2：

**输入：**matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

提示：

m == matrix.length

n == matrix[i].length

1 <= m, n <= 10

-100 <= matrix[i][j] <= 100

维护四个边界，每次到达边界则转换方向再收缩即可

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class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>> &matrix) {
        vector<pair<int, int>> dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int flag = 0, top = 0, left = 0;
        int down = matrix.size() - 1;
        int right = matrix[0].size() - 1;
        vector<int> ans;
        int i = 0, j = -1;
        while (top <= down && left <= right) {
            i += dir[flag].first;
            j += dir[flag].second;
            ans.push_back(matrix[i][j]);
            if (j == right && flag == 0) {
                top++;
                flag = 1;
            } else if (i == down && flag == 1) {
                right--;
                flag = 2;
            } else if (j == left && flag == 2) {
                down--;
                flag = 3;
            } else if (i == top && flag == 3) {
                left++;
                flag = 0;
            }
        }
        return ans;
    }
};