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.
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null)
return 0;
int m = obstacleGrid.length, n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
// Check first element
if (obstacleGrid[0][0] == 1)
return 0;
else
dp[0][0] = 1;
// Left column
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 1)
dp[i][0] = 0;
else
dp[i][0] = dp[i - 1][0];
}
// Top row
for (int i = 1; i < n; i++) {
if (obstacleGrid[0][i] == 1)
dp[0][i] = 0;
else
dp[0][i] = dp[0][i - 1];
}
// Inside
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1)
dp[i][j] = 0;
else
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
}