A robot is located at the top-left corner of a m x n grid. The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid. 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.

Note: m and n will be at most 100. Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length == 0) return -1;
        if(obstacleGrid[0] == null || obstacleGrid[0].length == 0) return -1;

        int row = obstacleGrid.length;
        int col = obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1 || obstacleGrid[row-1][col-1] == 1) return 0;

        int[][] dp = new int[row][col];

        //initialize
        dp[0][0] = 1;
        for(int i = 0; i < col; i++) {
            if(obstacleGrid[0][i] == 1) break;
            dp[0][i] = 1;
        }
        for(int i = 0; i < row; i++) {
            if(obstacleGrid[i][0] == 1) break;
            dp[i][0] = 1;
        }

        //iterate
        for(int i = 1; i < row; i++) {
            for(int j = 1; j < col; j++) {
                if(obstacleGrid[i][j] == 1) dp[i][j] = 0;
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }

        return dp[row-1][col-1];
    }
}

简化了一些conor case
class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length == 0) return -1;
        if(obstacleGrid[0] == null || obstacleGrid[0].length == 0) return -1;

        int row = obstacleGrid.length;
        int col = obstacleGrid[0].length;

        int[][] dp = new int[row][col];

        //initialize
        for(int i = 0; i < col; i++) {
            if(obstacleGrid[0][i] == 1) break;
            dp[0][i] = 1;
        }
        for(int i = 0; i < row; i++) {
            if(obstacleGrid[i][0] == 1) break;
            dp[i][0] = 1;
        }

        //iterate
        for(int i = 1; i < row; i++) {
            for(int j = 1; j < col; j++) {
                if(obstacleGrid[i][j] == 1) dp[i][j] = 0;
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }

        return dp[row-1][col-1];
    }
}