题目描述：

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]

 逻辑：

沿着某一边不断访问数组元素，每访问结束一边，就将代表该条边的变量往中心靠拢一次。当访问最底边或者最左边时，进行判断，是否发生触底事件。

class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {vector<int> result;if (matrix.empty()) return result;int top = 0, bottom = matrix.size() - 1;int left = 0, right = matrix[0].size() - 1;while (top <= bottom && left <= right) {// Traverse from left to right along the top rowfor (int i = left; i <= right; ++i) {result.push_back(matrix[top][i]);}++top;  // Move down to the next row// Traverse from top to bottom along the right columnfor (int i = top; i <= bottom; ++i) {result.push_back(matrix[i][right]);}--right;  // Move left to the next columnif (top <= bottom) {// Traverse from right to left along the bottom rowfor (int i = right; i >= left; --i) {result.push_back(matrix[bottom][i]);}--bottom;  // Move up to the previous row}if (left <= right) {// Traverse from bottom to top along the left columnfor (int i = bottom; i >= top; --i) {result.push_back(matrix[i][left]);}++left;  // Move right to the next column}}return result;}
};