[leetcode]54.螺旋矩阵

螺旋矩阵
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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

思路：
初始化定义四条边界，然后顺时针走完一圈后边界向内收缩，直至边界重合。

AC代码：

class Solution {
   public:
    vector<int> spiralOrder(vector<vector<int>> &matrix) {
        vector<int> ans;
        if (matrix.empty() || matrix[0].empty()) return {};
        int height = matrix.size(), width = matrix[0].size();
        int up = 0, down = height - 1, left = 0, right = width - 1;
        while (true) {
            for (int i = left; i <= right; i++) {  //从左至右
                ans.push_back(matrix[up][i]);
            }
            if (++up > down) break;
            for (int j = up; j <= down; j++) {  //从上至下
                ans.push_back(matrix[j][right]);
            }
            if (--right < left) break;
            for (int i = right; i >= left; i--) {
                ans.push_back(matrix[down][i]);
            }
            if (--down < up) break;
            for (int j = down; j >= up; j--) {
                ans.push_back(matrix[j][left]);
            }
            if (++left > right) break;
        }
        return ans;
    }
};