LC54. 螺旋矩阵

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：

img

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：

img

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

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int left = 0, right = n - 1, top = 0, low = m - 1;
        vector<int> res;
        while(res.size() < n * m) {
            if(top <= low) {
                for(int i = left; i <= right ; i++) {
                    res.push_back(matrix[top][i]);
                }
                top++;
            }
            if(left <= right) {
                for(int i = top; i <= low; i++) {
                    res.push_back(matrix[i][right]);
                }
                right--;
            }
            if(top <= low) {
                for(int i = right; i >= left; i--) {
                    res.push_back(matrix[low][i]);
                }
                low--;
            }
            if(left <= right) {
                for(int i = low; i >= top; i--) {
                    res.push_back(matrix[i][left]);
                }
                left++;
            }
        }
        
        return res;
    }
};