63. Unique Path II Leetcode Python

带障碍的唯一路径

最新推荐文章于 2025-12-02 18:04:47 发布

原创最新推荐文章于 2025-12-02 18:04:47 发布 · 925 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#动态规划 #DP #leetcode #python #Array

leetcode 专栏收录该内容

166 篇文章

订阅专栏

本文介绍了一种使用动态规划解决网格中带有障碍物的唯一路径数量的问题。文章提供了一个具体的Python实现示例，并解释了如何根据网格上的障碍物调整算法。

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.

Note: m and n will be at most 100.

这题的做法和前面的unique path是一样的，都是用的动态规划。不同之处在于需要判断有没有点为1.

直接上代码：

class Solution:
    # @param obstacleGrid, a list of lists of integers
    # @return an integer
    def uniquePathsWithObstacles(self, obstacleGrid):
        rown=len(obstacleGrid)
        coln=len(obstacleGrid[0])
        dp=[[0 for i in range(coln)] for j in range(rown)]
        if obstacleGrid[0][0]!=1:
            dp[0][0]=1
        for row in range(1,rown):
            if obstacleGrid[row][0]!=1 and dp[row-1][0]==1:
                dp[row][0]=1
            else:
                break
        for col in range(1,coln):
            if obstacleGrid[0][col]!=1 and dp[0][col-1]==1:
                dp[0][col]=1
            else:
                break
        for row in range(1,rown):
            for col in range(1,coln):
                if obstacleGrid[row][col]==1:
                    dp[row][col]=0
                else:
                    dp[row][col]=dp[row-1][col]+dp[row][col-1]
        return dp[rown-1][coln-1]