思路：

• dfs，超时

class Solution {public:    int uniquePathsWithObstacles(vector
    
     
      >& obstacleGrid) {        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        return dfs(obstacleGrid,m,n,0,0);    }    int dfs(vector
      
       
        >& obstacleGrid,int m,int n,int curi,int curj){       if(curi > m-1 || curj > n-1) return 0;       if(obstacleGrid[curi][curj] == 1) return 0;       else{           if(curi == m-1 && curj == n-1) return 1;           return dfs(obstacleGrid,m,n,curi+1,curj) + dfs(obstacleGrid,m,n,curi,curj+1);       }    }};
       
      
     
    

• DP。

class Solution {public:    int uniquePathsWithObstacles(vector
    
     
      >& obstacleGrid) {        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        vector
      
       
        >dp(m,vector
        
         (n,0));        if(obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1) return 0;        dp[0][0] = 1;        for(int i = 1; i < n; i++){            if(obstacleGrid[0][i] == 1) dp[0][i] = 0;            else{                dp[0][i] = dp[0][i-1];            }        }        for(int i = 1; i < m; i++){            if(obstacleGrid[i][0] == 1) dp[i][0] = 0;            else{                dp[i][0] = dp[i-1][0];            }        }                for(int i = 1; i < m; i++){            for(int j = 1; j < n; j++){                if(obstacleGrid[i][j] == 1) dp[i][j] = 0;                else dp[i][j] = dp[i-1][j] + dp[i][j-1];            }        }        return dp[m-1][n-1];            }};