代码随想录算法训练营day39|62.不同路径、63. 不同路径 II-63. 不同路径 II ​​​​​​​

时间：2024-03-12 18:00:06

https://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.htmlhttps://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.html

视频讲解：动态规划，这次遇到障碍了| LeetCode：63. 不同路径 II_哔哩哔哩_bilibili

与上一题整体思路大体一致。

注意点：

1.要知道遇到障碍dp[i][j]保持0就可以了。

2.初始化的部分，障碍及之后部分应该都是0

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        if obstacleGrid[m - 1][n - 1] == 1 or obstacleGrid[0][0] == 1:
            return 0

        dp = [[0]*n for _ in range(m)]

        #初始化
        for i in range(m):
            if obstacleGrid[i][0] == 0:
                dp[i][0] = 1
            else:
                break
        
        for j in range(n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = 1
            else:
                break
        
        for i in range(1,m):
            for j in range(1,n):
                if obstacleGrid[i][j] == 1:
                    continue
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
        
        return dp[m-1][n-1]

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