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[leetcode DP]63. Unique Paths II

2024-09-08 08:44:17   218人阅读

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.

障碍物对应的dp表中设为0即可，为了使代码简洁，多加了两行dp[i][0]和dp[0][j]

 1 class Solution(object):
 2     def uniquePathsWithObstacles(self, obstacleGrid):
 3         m,n = len(obstacleGrid),len(obstacleGrid[0])
 4         dp = [[0 for j in range(n+1)] for i in range(m+1)]
 5         dp[0][1] = 1
 6         for i in range(1,m+1):
 7             for j in range(1,n+1):
 8                 if not obstacleGrid[i-1][j-1]:
 9                     dp[i][j] = dp[i][j-1] + dp[i-1][j]
10         return dp[m][n]
11

 还有一种占用O(N)空间的方法，感觉挺不错的

思路：用一个数组tmp记录每一行中每一个位置拥有的次数，每到一个没有障碍物的位置，那么这个位置自动继承上一个位置上所拥有的次数

 1 class Solution(object):
 2     def uniquePathsWithObstacles(self, obstacleGrid):
 3         m,n=len(obstacleGrid),len(obstacleGrid[0])
 4         tmp = [0]*(n+1)
 5         tmp [n-1] = 1
 6         for i in range(m-1,-1,-1):
 7             for j in range(n-1,-1,-1):
 8                 if obstacleGrid[i][j]:
 9                     tmp[j] = 0
10                 else:
11                     tmp[j] += tmp [j+1]
12         return tmp[0]
13

 

[leetcode DP]63. Unique Paths II


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