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.
Note: m and n will be at most 100.
这题的做法和前面的unique path是一样的,都是用的动态规划。不同之处在于需要判断有没有点为1.
直接上代码:
class Solution: # @param obstacleGrid, a list of lists of integers # @return an integer def uniquePathsWithObstacles(self, obstacleGrid): rown=len(obstacleGrid) coln=len(obstacleGrid[0]) dp=[[0 for i in range(coln)] for j in range(rown)] if obstacleGrid[0][0]!=1: dp[0][0]=1 for row in range(1,rown): if obstacleGrid[row][0]!=1 and dp[row-1][0]==1: dp[row][0]=1 else: break for col in range(1,coln): if obstacleGrid[0][col]!=1 and dp[0][col-1]==1: dp[0][col]=1 else: break for row in range(1,rown): for col in range(1,coln): if obstacleGrid[row][col]==1: dp[row][col]=0 else: dp[row][col]=dp[row-1][col]+dp[row][col-1] return dp[rown-1][coln-1]
时间: 2024-11-04 13:53:08