Unique Paths II

 [解题思路]
和Unique Path一样的转移方程：
Step[i][j] = Step[i-1][j] + Step[i][j-1] if Array[i][j] ==0
or            = 0 if Array[i][j] =1


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