LeetCode 63: 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.
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.
Note: m and n will be at most 100.

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // DP problem
        // Number of unique paths at (m, n) S[m-1][n-1] = S[m-1][n-2] + S[m-2][n-1] if obstacleGrid[m-1][n-1] == 0;
        // Otherwise, S[m-1][n-1] = 0.
        int m = obstacleGrid.length;
        
        if (m == 0)
            return 0;
            
        int n = obstacleGrid[0].length;
        
        int[][] S = new int[m][n];
        
        // If the source is blocked, we don't need calculate it any more.
        if (obstacleGrid[0][0] == 0)
            S[0][0] = 1;
        else
            return 0;
        
        // Note: The first row only has one way.
        if (n > 1)
            for (int i = 1; i < n; i++)
                if (obstacleGrid[0][i] == 0)
                    S[0][i] = S[0][i-1];
                else
                    S[0][i] = 0;
        
        // Note: The first column only has one way.   
        if (m > 1)
            for (int i = 1; i < m; i++)
                if (obstacleGrid[i][0] == 0)
                    S[i][0] = S[i-1][0];
                else
                    S[i][0] = 0; 
        
        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
                if (obstacleGrid[i][j] == 0)
                    S[i][j] = S[i][j-1] + S[i-1][j];
                else
                    S[i][j] = 0;
                
        return S[m-1][n-1];        
    }
}