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.
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
int m=obstacleGrid.size();
if(m==0) return 0;
int n=obstacleGrid[0].size();
int path[n];
path[0]=1-obstacleGrid[0][0];
for(int i=1;i<n;i++)
if(obstacleGrid[0][i]==1) path[i]=0;
else path[i]=path[i-1];
for(int i=1;i<m;i++)
{
if(obstacleGrid[i][0]==1) path[0]=0;
for(int j=1;j<n;j++)
if(obstacleGrid[i][j]==1) path[j]=0;
else path[j]=path[j-1]+path[j];
}
return path[n-1];
}
};
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
int m=obstacleGrid.size();
if(m==0) return 0;
int n=obstacleGrid[0].size();
int path[n];
path[0]=1-obstacleGrid[0][0];
for(int i=1;i<n;i++)
if(obstacleGrid[0][i]==1) path[i]=0;
else path[i]=path[i-1];
for(int i=1;i<m;i++)
{
if(obstacleGrid[i][0]==1) path[0]=0;
for(int j=1;j<n;j++)
if(obstacleGrid[i][j]==1) path[j]=0;
else path[j]=path[j-1]+path[j];
}
return path[n-1];
}
};