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     }    

 

posted @ 2015-06-16 23:30  Rosanne  阅读(190)  评论(0编辑  收藏  举报