python第九周作业2：LeetCode Unique Paths II

最新推荐文章于 2024-07-26 10:47:39 发布

原创最新推荐文章于 2024-07-26 10:47:39 发布 · 169 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一个机器人在含有障碍物的网格中寻找从起点到终点所有可能路径的算法。该算法通过动态规划的方法来计算每一点可达路径的数量，并最终返回目标点的路径数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

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.

Note: m and n will be at most 100.

AC代码：

class Solution:   
    def uniquePathsWithObstacles(self, obstacleGrid):  
        m = len(obstacleGrid)  
        n = len(obstacleGrid[0])  
          
        if obstacleGrid[0][0] == 1:  
            return 0  
        elif m == 1 and n == 1:  
            return 1  
          
        paths = [[] for i in range(m)]   
          
        for i in range(m):  
            if obstacleGrid[i][0] == 1:  
                while(i < m):  
                    paths[i].append(0)  
                    i += 1  
                break;  
            else:  
                paths[i].append(1)  
          
        for j in range(1, n):  
            if obstacleGrid[0][j] == 1:  
                while(j < n):  
                    paths[0].append(0)  
                    j += 1  
                break;  
            else:  
                paths[0].append(1)  
                  
        for i in range(1, m):  
            for j in range(1, n):  
                if obstacleGrid[i][j] == 1:  
                    paths[i].append(0)  
                else:  
                    paths[i].append(paths[i][j - 1] + paths[i - 1][j])  
                      
        return paths[m - 1][n - 1]