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.
思路：这题理解题意之后并不比上一题麻烦，只是我在刚开始的时候，没有完全想对思路。
在给第一行和第一列赋值时，如果有障碍了，则后面的则全部赋值为0，表示通不过。
具体代码如下：
public class Solution {
public int uniquePathsWithObstacles(int[][] f) {
if(f.length == 0 || f[0].length == 0 || f[0][0] > 0)
return 0;
//其余情况
for(int i = 0; i < f.length; i++){
if(f[i][0] > 0){//如果有障碍，后面全部为0
while(i < f.length){
f[i++][0] = 0;
}
}else{
f[i][0] = 1;//没有障碍赋值为1
}
}
for(int i = 1; i < f[0].length; i++){//从i=1开始，因为f[0][0]值已经改变
if(f[0][i] > 0){//如果有障碍，后面全部为0
while(i < f[0].length){
f[0][i++] = 0;
}
}else{
f[0][i] = 1;
}
}
//循环走到最终点
for(int i = 1; i < f.length; i++)
for(int j = 1; j < f[0].length; j++){
if(f[i][j] > 0)
f[i][j] = 0;//有障碍，设置0，表示不通
else
f[i][j] = f[i-1][j] + f[i][j-1];
}
return f[f.length-1][f[0].length-1];
}
}