LeetCode063. Unique Paths II-优快云博客

本文链接：https://blog.youkuaiyun.com/AIpush/article/details/102790160

本文探讨了在一个由障碍物和空地组成的网格中，机器人从左上角到右下角的不同路径数量计算方法。考虑到起点可能为障碍物，以及障碍物对路径的影响，文章提供了两种实现路径计数的算法。

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

Note: m and n will be at most 100.

Example 1:

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

这道题容易少考虑几种形式:
1.起点就是障碍

2.当第一行或者第一列某一点存在障碍,那么这一行或一列之后都是无法到达.

3.本题最初用1表示障碍,0表示可行,但在计算时候我们一般会反过来.这时候只需用1 - obstacleGrid[i][j]表达即可.

class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        if obstacleGrid[0][0] == 1: return 0
        r,c = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[1 for _ in range(c)] for _ in range(r)]
        for i in range(1, r):
            dp[i][0] = dp[i - 1][0] * (1 - obstacleGrid[i][0])
        for i in range(1, c):
            dp[0][i] = dp[0][i - 1] * (1 - obstacleGrid[0][i])

        for i in range(1, r):
            for j in range(1, c):
                dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) * (1 - obstacleGrid[i][j])
        return(dp[-1][-1])

还有更为简洁的in place算法

def uniquePathsWithObstacles(self, obstacleGrid):
    if not obstacleGrid:
        return 
    r, c = len(obstacleGrid), len(obstacleGrid[0])
    obstacleGrid[0][0] = 1 - obstacleGrid[0][0]
    for i in xrange(1, r):
        obstacleGrid[i][0] = obstacleGrid[i-1][0] * (1 - obstacleGrid[i][0])
    for i in xrange(1, c):
        obstacleGrid[0][i] = obstacleGrid[0][i-1] * (1 - obstacleGrid[0][i])
    for i in xrange(1, r):
        for j in xrange(1, c):
            obstacleGrid[i][j] = (obstacleGrid[i-1][j] + obstacleGrid[i][j-1]) * (1 - obstacleGrid[i][j])
    return obstacleGrid[-1][-1]