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 }