LintCode-Unique Path II

標籤：

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 }

 

LintCode-Unique Path II