lintcode-easy-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 0respectively 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.

public class Solution {
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // write your code here
        
        if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0] == null || obstacleGrid[0].length == 0)
            return 0;
        
        if(obstacleGrid[0][0] == 1)
            return 0;
        
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] result = new int[m][n];
        
        for(int i = 0; i < m; i++){
            if(obstacleGrid[i][0] == 1){
                while(i < m){
                    result[i][0] = 0;
                    i++;
                }
            }
            else
                result[i][0] = 1;
        }
        
        for(int i = 0; i < n; i++){
            if(obstacleGrid[0][i] == 1){
                while(i < n){
                    result[0][i] = 0;
                    i++;
                }
            }
            else
                result[0][i] = 1;
        }
        
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                if(obstacleGrid[i][j] == 0)
                    result[i][j] = result[i - 1][j] + result[i][j - 1];
                else    
                    result[i][j] = 0;
            }
        }
        
        return result[m - 1][n - 1];
    }
}