update 63. 不同路径 II：提供一维dp数组的Python解法

problems/0063.不同路径II.md

@@ -232,6 +232,38 @@ class Solution:
         return dp[-1][-1]
 ```
 
+```python
+class Solution:
+    """
+    使用一维dp数组
+    """
+
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        m, n = len(obstacleGrid), len(obstacleGrid[0])
+
+        # 初始化dp数组
+        # 该数组缓存当前行
+        curr = [0] * n
+        for j in range(n):
+            if obstacleGrid[0][j] == 1:
+                break
+            curr[j] = 1
+            
+        for i in range(1, m): # 从第二行开始
+            for j in range(n): # 从第一列开始，因为第一列可能有障碍物
+                # 有障碍物处无法通行，状态就设成0
+                if obstacleGrid[i][j] == 1:
+                    curr[j] = 0
+                elif j > 0:
+                    # 等价于
+                    # dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
+                    curr[j] = curr[j] + curr[j - 1]
+                # 隐含的状态更新
+                # dp[i][0] = dp[i - 1][0]
+        
+        return curr[n - 1]
+```
+
 
 Go：