动态规划--unique paths 2

1、题目：

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

2、解答：这个比上个问题，多了一些不可达的点。也就是边界条件多了一些。

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3、代码

C++代码

    

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
       int m = obstacleGrid.size();
       int n = obstacleGrid[0].size();
       vector<vector<int> > dp(m+1,vector<int>(n+1,0));
       //dp[0][1] = 1 设置为1 是为了统一边界。即走第一个格子的时候，为1，这个设置，自己好好琢磨
       //dp[0][1] = 1 设置这两个中的一个为1
        dp[1][0] = 1;
        
       for(int i=1;i<=m;i++)
           for(int j=1;j<=n;j++)
               if(!obstacleGrid[i-1][j-1])
                   dp[i][j] = dp[i-1][j] + dp[i][j-1];
       return dp[m][n];
    }
};

python代码

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