Lintcode-unique Path II

Follow up for "Unique Paths":

Now consider if some obstacles is added to the grids. How many unique paths would there be?

An obstacle and empty space are marked as and respectively in the 1 0 grid.

Note

m and N would be is at most 100.

Example

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 .

Analysis:

DP:D[I][J] = D[i][j-1]+d[i-1][j].

Note:we can use 1D array to perform the DP. Since D[i][j] depends on d[i][j-1], i.e., the new d[][j-1], we should increase J from 0 to end. If D[i][j] depends on d[i-1][j-1] and we should decrease J from end to 0.

Solution:

1  Public classSolution {2     /**3      * @paramobstaclegrid:a List of lists of integers4      * @return: An integer5      */6      Public intUniquepathswithobstacles (int[] obstaclegrid) {7         intRowNum =obstaclegrid.length;8         if(rownum==0)return0;9         intColnum = obstaclegrid[0].length;Ten         if(colnum==0)return0; One         if(obstaclegrid[0][0]==1)return0;  A  -         int[] Path =New int[Colnum]; -Path[0] =1; the          for(inti=1;i<colnum;i++) -             if(obstaclegrid[0][i]==1) path[i]=0; -             ElsePath[i] = path[i-1]; -  +          for(inti=1;i<rownum;i++){ -             if(obstaclegrid[i][0]==1) path[0]=0; +              for(intj=1;j<colnum;j++) A                 if(obstaclegrid[i][j]==1) path[j]=0; at                 Elsepath[j]=path[j-1]+Path[j]; -              -         } -              -  -         returnPath[colnum-1]; in     } -}

Lintcode-unique Path II