LintCode-Unique Path II

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.

Note

m and n will be 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] then we should decrease j from end to 0.

Solution:

 1 public class Solution {
 2     /**
 3      * @param obstacleGrid: A list of lists of integers
 4      * @return: An integer
 5      */
 6     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 7         int rowNum = obstacleGrid.length;
 8         if (rowNum==0) return 0;
 9         int colNum = obstacleGrid[0].length;
10         if (colNum==0) return 0;
11         if (obstacleGrid[0][0]==1) return 0;
12
13         int[] path = new int[colNum];
14         path[0] =1;
15         for (int i=1;i<colNum;i++)
16             if (obstacleGrid[0][i]==1) path[i]=0;
17             else path[i] = path[i-1];
18
19         for (int i=1;i<rowNum;i++){
20             if (obstacleGrid[i][0]==1) path[0]=0;
21             for (int j=1;j<colNum;j++)
22                 if (obstacleGrid[i][j]==1) path[j]=0;
23                 else path[j]=path[j-1]+path[j];
24
25         }
26
27
28         return path[colNum-1];
29     }
30 }