Leetcode题集——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 as1and0respectively 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 is2.

Note: m and n will be at most 100.
递推式还是跟 Unique
Paths 一样，只是每次我们要判断一下是不是障碍，如果是，则res[i][j]=0，从该节点之后的所有节点为0，否则还是res[i][j]=res[i-1][j]+res[i][j-1]。实现中还是只需要一个一维数组，因为更新的时候所需要的信息足够了。这样空间复杂度是是O(n).

int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
int m=obstacleGrid.size();
int n=obstacleGrid[0].size();

//鲁棒性检查
if(obstacleGrid.empty()||m==0||n==0)
return 0;

vector<int> v(n);
v[0]=1;

for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
if(obstacleGrid[i][j]==0)
{
if(j>0)
v[j]=v[j]+v[j-1];
}
else
v[j]=0;
}
return v[n-1];
}