A four digit number is called a doublet if any of its digit is the same as only one neighbour. (For exampel, 1221 is doublets but 1222 is not), Number of such doublets are

Case–1 abbc
We can choose a in 9 ways, b in 9 ways, and c in 9 ways. Hence, there are $9^3$ possibilities for this case.

Case–2 bbac
There are 9 possibilities for b, and 9 possibilities for each a and c. Once again, we have that there are $9^3$ possibilities for this case.

Case – 3 acbb
Once again, there are 9 possibilties for a and c, and 9 possibilties for b. hence, these are $9^3$ possibilities for this case.
Case – 4 aabb
We have 9 possibilities for a and 9 possibilities for b. We thus have 81 possibilities for this case. We can now sum up our answers, to get a total of

${3.9}^3+81=2268\text{ total ways }$