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update 63. 不同路径 II:提供一维dp数组的Python解法
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@ -232,6 +232,38 @@ class Solution:
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return dp[-1][-1]
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return dp[-1][-1]
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```
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```
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```python
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class Solution:
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"""
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使用一维dp数组
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"""
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def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
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m, n = len(obstacleGrid), len(obstacleGrid[0])
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# 初始化dp数组
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# 该数组缓存当前行
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curr = [0] * n
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for j in range(n):
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if obstacleGrid[0][j] == 1:
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break
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curr[j] = 1
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for i in range(1, m): # 从第二行开始
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for j in range(n): # 从第一列开始,因为第一列可能有障碍物
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# 有障碍物处无法通行,状态就设成0
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if obstacleGrid[i][j] == 1:
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curr[j] = 0
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elif j > 0:
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# 等价于
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# dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
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curr[j] = curr[j] + curr[j - 1]
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# 隐含的状态更新
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# dp[i][0] = dp[i - 1][0]
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return curr[n - 1]
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```
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Go:
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Go:
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