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.

解析：

本题相对于Unique Paths的难点在于矩阵的初始化，这里将数据为1所对应到dp[]][]矩阵中所对应位置的数据置0即可。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m=obstacleGrid.length;
        int n=obstacleGrid[0].length;
     
        if(m==0||n==0)
        return 0;
        int[][] dp=new int[m+1][n+1];
        if(obstacleGrid[0][0]!=1)
        {
        	dp[1][1]=1;
        }
            for(int i=2;i<=m;i++)
        {
            if(obstacleGrid[i-1][0]!=1&&dp[i-1][1]!=0)
            dp[i][1]=1;
        }
        for(int j=2;j<=n;j++)
        {
            if(obstacleGrid[0][j-1]!=1&&dp[1][j-1]!=0)
            dp[1][j]=1;
        }
        
        for(int i=2;i<=m;i++)
        {
            for(int j=2;j<=n;j++)
            {
                if(obstacleGrid[i-1][j-1]!=1)
                 dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m][n];
    }
}