leetcode——63. 不同路径 II
作者:互联网
class Solution: def uniquePathsWithObstacles(self, obstacleGrid) -> int: if len(obstacleGrid)<1: return 0 m=len(obstacleGrid) n=len(obstacleGrid[0]) if m==1 and n==1: if obstacleGrid[0][0]==0: return 1 else: return 0 memo=[[0 for i in range(n)] for i in range(m)]#定义空二维数组 for i in range(n): if obstacleGrid[0][i]==1: memo[0][i]=0 for k in range(i+1,n): memo[0][k]=0 break else: memo[0][i]=1 for j in range(m): if obstacleGrid[j][0]==1: memo[j:][0]=[0]*(m-j-1) break else: memo[j][0]=1 for i in range(1,m): for j in range(1,n): if obstacleGrid[i][j]==1: memo[i][j]=0 else: memo[i][j]=memo[i-1][j]+memo[i][j-1] return memo[m-1][n-1]执行用时 :64 ms, 在所有 python3 提交中击败了76.80%的用户 内存消耗 :13.7 MB, 在所有 python3 提交中击败了5.19%的用户 ——2019.10.14
标签:obstacleGrid,击败,用户,II,63,2019.10,提交,leetcode,python3 来源: https://www.cnblogs.com/taoyuxin/p/11673548.html