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.
Example
For example,
There is one obstacle in the middle of a 3×3 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.
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
// write your code here
if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0 || obstacleGrid[0][0] == 1) {
return 0;
}
int row = obstacleGrid.length;
int col = obstacleGrid[0].length;
int[][] path = new int[row][col];
path[0][0] = 1;
//init first col
for (int i = 1; i < row; i++) {
path[i][0] = obstacleGrid[i][0] == 1 ? 0 : path[i - 1][0];
}
//init first row
for (int j = 1; j < col; j++) {
path[0][j] = obstacleGrid[0][j] == 1 ? 0 : path[0][j - 1];
}
for (int i = 1; i < row; i++) {
for (int j = 1; j < col; j++) {
path[i][j] = obstacleGrid[i][j] == 1 ? 0 : path[i - 1][j] + path[i][j - 1];
}
}
return path[row - 1][col - 1];
}
