leetcode 63. Unique Paths II

最新推荐文章于 2024-07-26 10:47:39 发布

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本文探讨了在存在障碍物的网格中，机器人从左上角到右下角的路径规划问题。通过动态规划方法，详细解析了如何计算在3x3网格中，避开中间障碍物到达目标的路径数量。

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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

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

分析：
这道题是之前那道 Unique Paths的延伸，在路径中加了一些障碍物，还是用动态规划Dynamic Programming来解，不同的是当遇到为1的点，将该位置的dp数组中的值清零，其余和之前那道题并没有什么区别

def uniquePathsWithObstacles(self, obgrid):
    path = [0]*len(obgrid[0])
    if obgrid[0][0] == 1:
        return 0
     path[0]=1
     for i in range(len(obgrid)):
         for j in range(len(obgrid[0])):
             if obgrid[i][j] == 1:
                  path[j] = 0
             elif j > 0:
                 path[j] += path[j-1]
    return path[len(obgrid[0])-1]