Update 0054.螺旋矩阵.md

problems/0054.螺旋矩阵.md

@@ -251,6 +251,60 @@ var spiralOrder = function(matrix) {
     return arr
 };
 ```
+### Python
+
+```python
+class Solution(object):
+    def spiralOrder(self, matrix):
+        """
+        :type matrix: List[List[int]]
+        :rtype: List[int]
+        """
+        if len(matrix)==0 or len(matrix[0])==0 : # 判定List是否为空
+            return []
+        row, col = len(matrix), len(matrix[0]) # 行数，列数
+        loop = min(row, col) // 2 # 循环轮数
+        stx, sty = 0, 0 # 起始x，y坐标
+        i, j =0, 0
+        count = 0  # 计数
+        offset = 1  # 每轮减少的格子数
+        result = [0]*(row*col)
+        while loop>0 :# 左闭右开
+            i, j = stx, sty
+            while j < col - offset :   # 从左到右
+                result[count] = matrix[i][j]
+                count += 1
+                j += 1  
+            while i < row - offset : # 从上到下
+                result[count] = matrix[i][j]
+                count += 1
+                i += 1
+            while j>sty :  # 从右到左
+                result[count] = matrix[i][j]
+                count += 1
+                j -= 1
+            while i>stx :  # 从下到上
+                result[count] = matrix[i][j]
+                count += 1
+                i -= 1
+            stx += 1
+            sty += 1
+            offset += 1
+            loop -= 1
+        if min(row, col) % 2 == 1 :  # 判定是否需要填充多出来的一行
+            i = stx
+            if row < col :
+                while i < stx + col - row + 1 :
+                    result[count] = matrix[stx][i]
+                    count += 1
+                    i += 1
+            else :
+                while i < stx + row - col + 1 :
+                    result[count] = matrix[i][stx]
+                    count += 1
+                    i += 1
+        return result
+```