ADD:
DP Subject:100 DP formula
63. Unique Paths II
Medium
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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.
Note: m and n will be at most 100.
Example 1:
Input: [ [0,0,0], [0,1,0], [0,0,0] ] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right
BONUS:
1.for python for loop don't have actual meaning so even "_" this kinds of thing can also used in for loop
for _ in range(5): print(_)
dp = [[0] * n for _ in range(m)]
using this method to initialize the two-dimension array
2.Python strip() 方法用于移除字符串头尾指定的字符(默认为空格)
import sys class Solution(object): def uniquePathsWithObstacles(self, obstacleGrid): """ :type obstacleGrid: List[List[int]] :rtype: int """ m,n=len(obstacleGrid),len(obstacleGrid[0]) #attention:0*n first multiply n for m the rank easy confuse dp=[[0]*n for _ in range(m)] if obstacleGrid[0][0]==0: dp[0][0]=1 for i in range(m): for j in range(n): if obstacleGrid[i][j]==0: if i==j==0: continue dp[i][j]=dp[i-1][j]+dp[i][j-1] return dp[m-1][n-1] solu=Solution() #obstacleGrid=input() myList2=[ [0,0,0], [0,1,0], [0,0,0] ] print(solu.uniquePathsWithObstacles(myList2))
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