Unique Paths II 解答

Question

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.

Solution

Similar with "Unique Paths", there are two differences to be considered:

1. for dp[0][i] and dp[i][0], if there exists previous element which equals to 1, then the rest elements are all unreachable.

2. for dp[i][j], if it equals to 1, then it‘s unreachable.

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         if (obstacleGrid == null)
 4             return 0;
 5         int m = obstacleGrid.length, n = obstacleGrid[0].length;
 6         int[][] dp = new int[m][n];
 7         // Check first element
 8         if (obstacleGrid[0][0] == 1)
 9             return 0;
10         else
11             dp[0][0] = 1;
12         // Left column
13         for (int i = 1; i < m; i++) {
14             if (obstacleGrid[i][0] == 1)
15                 dp[i][0] = 0;
16             else
17                 dp[i][0] = dp[i - 1][0];
18         }
19         // Top row
20         for (int i = 1; i < n; i++) {
21             if (obstacleGrid[0][i] == 1)
22                 dp[0][i] = 0;
23             else
24                 dp[0][i] = dp[0][i - 1];
25         }
26         // Inside
27         for (int i = 1; i < m; i++) {
28             for (int j = 1; j < n; j++) {
29                 if (obstacleGrid[i][j] == 1)
30                     dp[i][j] = 0;
31                 else
32                     dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
33             }
34         }
35         return dp[m - 1][n - 1];
36     }
37 }