COMPLETED GRID
REASONING
Any Skyscraper view with a 1 must have 9 on its edge, giving us four 9s along the edges, and lets us fill in all of the 9s. In the first column looking down, there are only two buildings seen, but the 8 cannot be on the other side of the 9, so it must be in the first row. Continuing in this column, the only remaining numbers to be placed are 1, 4, 6 and 7, and all but 7 are barred from row 2. In row 8, R8C1 cannot be 6, which is blocked. It also cannot be 4: there are four buildings seen, which must be R8C1-4. If R8C1 were 4, R8C2-3 would have to be between 4 and 9, but 5, 6, and 8 are all barred, leaving 7 the only possibility. Thus R8C1 must be 1.
Continuing in the lower left square, 7 must be in R8-9C3 by Sudoku rules, but it cannot be in R9 because of the 4 buildings seen up in column 3, so R8C3 is 7. Now the 6 in this square must be in R7C1-2, but if it were in R7C2, R7C1 would be 4, showing 4 buildings from the left. This lets us fill in the entire first column. The grid thus far:
Looking up top:
In the first row, 7 is blocked from the upper left square, and cannot be in the upper right square since we would then see three buildings from the right. So R1C5 is 7. The 2 building view looking down column 5 forces the 8 to be in R9C5, with Sudoku rules forcing R8C7 to be 8 as well. Back to the upper right square, its 8 has to be in R2-3C8, but it cannot be in row 3 since we would only see 3 buildings from the right. Sudoku rules let us finish row 2.
Some miscellaneous deductions come now. In row 3, since we only see two buildings from the left, R3C2 has to be 1. In column 4, the only place a 7 can go is in R9. In the top row, the two missing cells in R1 are 3 and 4, meaning the missing cells in the upper right square are 1 and 2, and their positions are forced. Column 5 only needs 1, 4 and 6, and their positions are forced too. The grid thus far:
At the bottom:
The 6 in the lower middle square is forced to be in R9C6. This leaves the possible values for R9C8 as 1, 3 and 4, which are all less than 6; thus the 7 in column 8 must be in R3C7 to see only three buildings from the bottom. This forces the 7 in row 7 to be in R7C9 by Sudoku rules, and R5C7 finishes the placement of the 7s.
Looking in row 5 from the right, to have 5 buildings seen we are now forced to have R5C6 to be 8, which lets us finish the placement of the 8s in R6C2 and R3C4, the latter of which forces R3C6 to be 4. Sudoku logic in column 6 forces R4C6 to be 3. The grid thus far:
Finishing up:
By Sudoku rules in column 2, the 6 must be in R5C2, the placement of which forces the entire middle left square. In the bottom row we cannot have 1 in R9C9, since we would see 5 buildings from the right, so R9C8 must be 1. We complete the placement of 1s with R5C9. This forces R6C9 to be 2, which in turn forces R4C4 to be 2. The rest is just Sudoku fill-ins.
