Merge pull request #1039 from leeeeeeewii/54

添加0054.螺旋矩阵 python版本

problems/0054.螺旋矩阵.md

@@ -131,7 +131,46 @@ public:
 * [59.螺旋矩阵II](https://leetcode-cn.com/problems/spiral-matrix-ii/)
 * [剑指Offer 29.顺时针打印矩阵](https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/)
 
+## 其他语言版本
+Python:
+```python
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        m, n = len(matrix), len(matrix[0])
+        left, right, up, down = 0, n - 1, 0, m - 1 # 定位四个方向的边界，闭区间
+        res = []
+        
+        while True:
+            for i in range(left, right + 1): # 上边，从左到右
+                res.append(matrix[up][i])
+            up += 1 # 上边界下移
 
+            if len(res) >= m * n: # 判断是否已经遍历完
+                break
+            
+            for i in range(up, down + 1): # 右边，从上到下
+                res.append(matrix[i][right])
+            right -= 1 # 右边界左移
+
+            if len(res) >= m * n:
+                break
+            
+            for i in range(right, left - 1, -1): # 下边，从右到左
+                res.append(matrix[down][i])
+            down -= 1 # 下边界上移
+
+            if len(res) >= m * n:
+                break
+
+            for i in range(down, up - 1, -1): # 左边，从下到上
+                res.append(matrix[i][left])
+            left += 1 # 左边界右移
+            
+            if len(res) >= m * n:
+                break
+
+        return res
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>