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 1
        return 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]*n
        for 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 0
        dp=[0]*len(obstacleGrid[0])
        for j in range(len(dp)):
            if obstacleGrid[0][j]:dp[j]=0
            elif j==0:dp[j]=1
            else: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]=0
                elif j!=0:
                    dp[j]+=dp[j-1]
        return dp[-1]