# Unique Path II

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.length == 0 || obstacleGrid == 1) {
return 0;
}
int row = obstacleGrid.length;
int col = obstacleGrid.length;
int[][] path = new int[row][col];
path = 1;
//init first col
for (int i = 1; i < row; i++) {
path[i] = obstacleGrid[i] == 1 ? 0 : path[i - 1];
}
//init first row
for (int j = 1; j < col; j++) {
path[j] = obstacleGrid[j] == 1 ? 0 : path[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];
}```