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]; }