62. 不同路径
递归栈很酷 但超时

class Solution:def uniquePaths(self, m: int, n: int) -> int:if m==1 or n==1:return 1return self.uniquePaths(m-1,n)+self.uniquePaths(m,n-1)

逐行dp

class Solution:def uniquePaths(self, m: int, n: int) -> int:dp=[1]*nfor j in range(1,m):for i in range(1,n):dp[i]+=dp[i-1]return dp[n-1]

63. 不同路径 II

j从0开始 每个都要判断

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