【leetcode】Array——Unique Paths II（63）

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.
因为刚学习了动态规划 所以这题容易了许多
用二维数组实现DP
陷阱：

第一步，不能把全部的第一行以及第一列全部初始化为1了，在“障碍”后面的，仍为零
如果终点是“障碍”，要单独考虑

代码：

public int uniquePathsWithObstacles(int[][]
obstacleGrid) {
int [][] a =
new int [obstacleGrid.length][obstacleGrid[0].length];
for(int
i=0;i<obstacleGrid.length;i++){
for(int
j=0;j<obstacleGrid[0].length;j++){
if(obstacleGrid[i][j]==0){
a[i][j]=0;
}else
if(obstacleGrid[i][j]==1){
a[i][j]=-1;
}
}
}

//dp
init
for(int
i=0;i<a.length;i++){
if(a[i][0]!=-1){
a[i][0]=1;
}else
break;
}
for(int
i=0;i<a[0].length;i++){
if(a[0][i]!=-1){
a[0][i]=1;
}else
break;
}

//start dp
for(int
i=1;i<a.length;i++){
for(int
j=1;j<a[0].length;j++){
if(a[i][j]==-1)
continue;
if(a[i-1][j]!=-1){
a[i][j]+=a[i-1][j];
}
if(a[i][j-1]!=-1){
a[i][j]+=a[i][j-1];
}
}
}

if(a[a.length-1][a[0].length-1]==-1)
return 0;

return
a[a.length-1][a[0].length-1];
}