##题目
####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.
##解题思路
该题是Unique Paths的扩展，思路类似，只是这边要处理障碍。动态规划的思路可以见那题。
##算法代码
代码采用JAVA实现：
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public class {
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++)
for(int j=0;j<n;j++)
dp[i][j]=0;
for(int i=0;i<n;i++)
{
if(obstacleGrid[0][i]!=1)
dp[0][i]=1;
else
break;
}
for(int j=0;j<m;j++)
{
if(obstacleGrid[j][0]!=1)
dp[j][0]=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]=0;
else
dp[i][j]=dp[i-1][j]+dp[i][j-1];
}
return dp[m-1][n-1];
}
}
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