Leetcode 63.不同路径II

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63. Unique Paths II

题目

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？

网格中的障碍物和空位置分别用 1 和 0 来表示。

说明：m 和 n 的值均不超过 100。

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示例 1:
输入:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
输出: 2
解释:
3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径：
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右

方法

机器人只能向下或者向右移动，因此第一行的cell只能左边到达，第一列的cell只能从上方到达，其他任意的cell可以从左边或者上方到达。如果该cell有阻碍物，则对任何路径都没有贡献

方法1：动态规划 & not in place

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class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        if not obstacleGrid:
            return
        r,c=len(obstacleGrid),len(obstacleGrid[0])
        dp=[[0 for _ in range(c)] for _ in range(r)]
        dp[0][0]=1-obstacleGrid[0][0]

        for i in range(1,r):
            dp[i][0]=dp[i-1][0]*(1-obstacleGrid[i][0])
        for i in range(1,c):
            dp[0][i]=dp[0][i-1]*(1-obstacleGrid[0][i])
        for i in range(1,r):
            for j in range(1,c):
                dp[i][j]=(dp[i-1][j]+dp[i][j-1])*(1-obstacleGrid[i][j])
        return dp[-1][-1]
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class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        if not obstacleGrid:
            return 0
        m,n=len(obstacleGrid),len(obstacleGrid[0])
        dp=[[0 for _ in range(n)] for _ in range(m)]
        dp[0][0]=1 if not obstacleGrid[0][0] else  0                        # 确定（0，0）的状态
        for i in range(1,m):                                                # 确定 除（0，0）外，第一列的状态
            dp[i][0]=1 if dp[i-1][0] and not obstacleGrid[i][0] else 0
        for i in range(1,n):                                                # 确定除（0，0）外，第一行的状态
            dp[0][i]=1 if dp[0][i-1] and not obstacleGrid[0][i] else 0
        for i in range(1,m):
            for j in range(1,n):
                if obstacleGrid[i][j]:
                    dp[i][j]=0
                else:
                    dp[i][j]=dp[i-1][j]+dp[i][j-1]
        return dp[-1][-1]
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class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])

        # If the starting cell has an obstacle, then simply return as there would be
        # no paths to the destination.
        if obstacleGrid[0][0] == 1:
            return 0

        # Number of ways of reaching the starting cell = 1.
        obstacleGrid[0][0] = 1

        # Filling the values for the first column
        for i in range(1,m):
            obstacleGrid[i][0] = int(obstacleGrid[i][0] == 0 and obstacleGrid[i-1][0] == 1)

        # Filling the values for the first row        
        for j in range(1, n):
            obstacleGrid[0][j] = int(obstacleGrid[0][j] == 0 and obstacleGrid[0][j-1] == 1)

        # Starting from cell(1,1) fill up the values
        # No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
        # i.e. From above and left.
        for i in range(1,m):
            for j in range(1,n):
                if obstacleGrid[i][j] == 0:
                    obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
                else:
                    obstacleGrid[i][j] = 0

        # Return value stored in rightmost bottommost cell. That is the destination.            
        return obstacleGrid[m-1][n-1]
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class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        r,c=len(obstacleGrid),len(obstacleGrid[0])
        dp=[[0 for _ in range(c)] for _ in range(r)]

        if obstacleGrid[0][0]==0:
            dp[0][0]=1

        for i in range(r):
            for j in range(c):
                if obstacleGrid[i][j]==0:
                    if i==j==0:
                        continue
                    else:
                        dp[i][j]=dp[i-1][j]+dp[i][j-1]
        return dp[-1][-1]
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class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        r,c=len(obstacleGrid),len(obstacleGrid[0])
        dp=[[0 for _ in range(c)] for _ in range(r)]

        if obstacleGrid[0][0]==0:
            dp[0][0]=1

        for i in range(r):
            for j in range(c):
                if obstacleGrid[i][j]==1:
                    dp[i][j]=0
                else:
                    if i!=0:
                        dp[i][j]+=dp[i-1][j]
                    if j!=0:
                        dp[i][j]+=dp[i][j-1]
        return dp[r-1][c-1]

方法2：动态规划 & in place

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class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        if not obstacleGrid:
            return
        r,c=len(obstacleGrid),len(obstacleGrid[0])
        obstacleGrid[0][0]=1-obstacleGrid[0][0]

        for i in range(1,r):
            obstacleGrid[i][0]=obstacleGrid[i-1][0]*(1-obstacleGrid[i][0])
        for i in range(1,c):
            obstacleGrid[0][i]=obstacleGrid[0][i-1]*(1-obstacleGrid[0][i])
        for i in range(1,r):
            for j in range(1,c):
                obstacleGrid[i][j]=(obstacleGrid[i-1][j]+obstacleGrid[i][j-1])*(1-obstacleGrid[i][j])
        return obstacleGrid[-1][-1]

参考资料