题目：

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.

题意：

紧跟着题目《Unique Paths》，现给出这样一题目：

假设在格子中加入一些障碍，会出现多少存在且唯一的不同路径呢？

障碍和空白格子分别被标记为1 and 0 .

比方一个3x3的格子中的中间存在一个障碍，例如以下所看到的：

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

算法分析：

     思路与题目《Unique Paths》类似，不同之处为：

     初始化边界上行和列时，出现障碍。后面路径数dp的都是0

     中间的格子出现障碍时，该格子dp表示的路径数直接填0

AC代码：

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