LeetCode 63. Unique Paths II Python Solution

最新推荐文章于 2024-03-21 18:04:53 发布

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本文介绍了一种使用动态规划解决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数量的方法。通过构建一个与原始网格相同大小的矩阵，并逐个计算每个单元格的最优解来实现。

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题目：

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

这道题是一个典型的动态规划，对我来说也比较陌生，也相对系统的稍微学习和了解了一下其算法。做法是先构建一个同样大小的矩阵，然后算每一个方块的最优解。

代码：

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        dp = [[0]*len(obstacleGrid[0]) for _ in range(len(obstacleGrid))]
        dp[0][0] = 1 if obstacleGrid[0][0] == 0 else 0
        for i in range(len(obstacleGrid)):
            for j in range(len(obstacleGrid[0])):
                if obstacleGrid[i][j] == 1:
                    dp[i][j] = 0
                else:
                    if i-1>=0 and j-1>=0:
                        dp[i][j] = dp[i-1][j] + dp[i][j-1]
                    else:
                        if j-1>=0:
                            dp[i][j] = dp[i][j-1]
                        if i-1>=0:
                            dp[i][j] = dp[i-1][j]
        return dp[-1][-1]