Unique Paths II

Description：

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.

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.

Note: m and n will be at most 100.

Code:

 1     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {

 2         int m = obstacleGrid.size();

 3         int n = obstacleGrid[0].size();

 4 

 5     //path[i][j]表示(0,0)号元素到(i,j)号元素的路径数，所以最终结果为path[m-1][j-1]

 6 

 7         int path[MAX][MAX] = {0};

 8         path[0][0] = (obstacleGrid[0][0] == 1)?0:1;

 9 

10         for (int i = 1; i < m; ++i)

11         {

12             path[i][0] = (obstacleGrid[i][0]==1)?0:path[i-1][0];

13         }

14 

15         for (int i = 1; i < n; ++i)

16         {

17             path[0][i] = (obstacleGrid[0][i]==1)?0:path[0][i-1];

18         }

19         

20         for (int i = 1; i < m; ++i)

21         {

22             for (int j = 1; j < n; ++j)

23             {

24                 path[i][j] = (obstacleGrid[i][j] == 1)?0:(path[i-1][j] + path[i][j-1]);

25             }

26         }

27         return path[m-1][n-1];

28     }