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  • 63.Unique Path II

    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?

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    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
    class Solution:
        def uniquePathsWithObstacles(self, obstacleGrid):
            """
            :type obstacleGrid: List[List[int]]
            :rtype: int
            """
            if obstacleGrid[0][0]==1:
                return 0
            m,n = len(obstacleGrid),len(obstacleGrid[0])
            dp = [[0 for i in range(n+2)] for j in range(m+2)]
            dp[1][2] = dp[2][1] =1
            for i in range(1,m+1):
                for j in range(1,n+1):
                    if obstacleGrid[i-1][j-1]==0:
                        if (i==1 and j==2) or (i==2 and j==1) or(i==1 and j==1):
                            dp[i][j]=1
                            continue
                        dp[i][j] = dp[i-1][j] + dp[i][j-1]
                    else:
                        dp[i][j]=0
            return dp[m][n]
    

    Easy question,if the position is obstacle,just set it zero.

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  • 原文地址:https://www.cnblogs.com/bernieloveslife/p/9762802.html
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