leetcode63. Unique Paths II

题目描述（难度：Medium）：

Follow up for "Unique Paths":
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.


For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
The total number of unique paths is 2.
Note: m and n will be at most 100.


Tags： Array， Dynamic Programming
Similar Problems： (M) Unique Paths

分析：求解path的种类数量（种类数动规/棋盘类动规）
        设dp[i][j]表示从左上角到达grid[i][j]时的种类数，
        if dp[i][j]==1, 则dp[i][j] = 0(不通达)
        if dp[i][j]==0, 则dp[i][j] = dp[i-1][j]+dp[i][j-1]

代码实现：

class Solution(object):    def uniquePathsWithObstacles(self, obstacleGrid):        """        :type obstacleGrid: List[List[int]]        :rtype: int        """        #1.先做输入处理        m = len(obstacleGrid)        n = len(obstacleGrid[0]) if m else 0        if m==0 or n==0:            return 0        #2.初始化，同时注意dp[0][0]的值        dp = [ [0]*n for i in range(m) ]        if obstacleGrid[0][0] == 0:            dp[0][0] = 1        for i in range(1, m):            if obstacleGrid[i][0] == 0:                dp[i][0] = dp[i-1][0]        for j in range(1, n):            if obstacleGrid[0][j] == 0:                dp[0][j] = dp[0][j-1]        #3.dp        for i in range(1, m):            for j in range(1, n):                if obstacleGrid[i][j] == 0:                    dp[i][j] = dp[i-1][j]+dp[i][j-1]        return dp[m-1][n-1]