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.
题解：和http://www.cnblogs.com/sunshineatnoon/p/3798167.html非常相似，要注意两点：
- 有障碍的地方能走到的方法数目是0；
- 初始化结果矩阵的时候，如果第一行（列）的某个位置有障碍，那么这一行（列）该元素后面所有的元素都是0，不能置为1；比如给定障碍矩阵[1,0,0]，那么初始化以后的矩阵应该为[0,0,0]，而不是[0,1,1]，因为(0,0)处有障碍，那么该矩阵的任何位置都是无法到达的。
代码如下：
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[][] PathNum = new int[m][n];
for(int i = 0;i < m;i++)
if(obstacleGrid[i][0] != 1)
PathNum[i][0] = 1;
else
break;
for(int i = 0;i < n;i++)
if(obstacleGrid[0][i] != 1)
PathNum[0][i] = 1;
else
break;
for(int i = 1;i < m;i++)
for(int j = 1;j < n;j++){
if(obstacleGrid[i][j] != 1)
PathNum[i][j] = PathNum[i-1][j]+PathNum[i][j-1];
}
return PathNum[m-1][n-1];
}
}