Unique Paths 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.

Notice: m and n will be at most 100.

Notes:

1. Coordinate DP

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0
        || obstacleGrid[0] == null || obstacleGrid[0].length == 0) {
            return 0;
        }

        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;

        // state
        int[][] f = new int[m][n];

        // initialize
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }

        f[0][0] = 1;
        for (int i = 1; i < m; ++i) {
            if (f[i-1][0] == 1 && obstacleGrid[i][0] == 0) {
                f[i][0] = 1;
            } else {
                f[i][0] = 0;
            }
        }

        for (int i = 1; i < n; ++i) {
            if (f[0][i-1] == 1 && obstacleGrid[0][i] == 0) {
                f[0][i] = 1;
            } else {
                f[0][i] = 0;
            }
        }

        // function
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                f[i][j] = obstacleGrid[i][j] == 1 ? 0 : f[i][j-1] + f[i-1][j];
            }
        }

        // answer
        return f[m-1][n-1];
    }