Sums are for heights of skyscrapers visible from the edges. Only those higher than the previous one(s) count toward the sums.
On the odd-numbered rows and columns, which are normal ones containing 1-9, each sum must include 9, the height of the highest skyscraper, which can be seen from each edge. However this doesn’t necessarily apply for the even-numbered rows and columns where repeated numbers are allowed; for these rows and columns the height of the highest skyscraper may be less than 9.
Since this an ORC puzzle, I’ve stated placements in odd rows/columns.
x not 0, y not 7, w is unknown.
The even-numbered have 5 three in a row, 1 four in a row, horizontally and/or vertically.
1. Skyscrapers starting with 8 or 9 must total 8 (possible for even rows/columns), 9 or 17. Since no totals are given as 17 and xy or 2w cannot be 8, 9 or 17, rows/columns starting/finishing with 8,9 must be ones with unspecified totals.
1a. Unspecified upper totals only in C79 -> R1C79 = {89}, locked for R1
1b. Unspecified left-hand totals only in R36 -> R36C1 = {89}, locked for C1
1c. Unspecified right-hand totals only in R15 -> R15C9 = {89}, locked for C9
1d. Lower total in C9 = 30 must contain both of 8,9 (cannot be [97.6.53] in R56789C9 because of NC while other sums containing 9 but not 8 contain 6 or 7 digits) -> R1C9 = 9, R1C7 = 8, placed for C7, R5C9 = 8, placed for R5
1e. Unspecified lower totals only in C247, no 8 in R9C7 -> 8 must be in one of R9C24, locked for R9, 9 must be in one of R9C247, locked for R9
Clean-ups:
R1C7 = 8 -> no 7 in R1C68, no 7,8,9 in R2C68, no 7,9 in R2C7
R5C9 = 8 -> no 7,8,9 in R46C8, no 7 in R46C9, no 7,9 in R5C8
R3C1 = {89} -> no 8,9 in R234C2
R6C1 = {89} -> no 9 in R5C2, no 8,9 in R7C2
8 in one of R9C24 -> no 7,8,9 in R8C3, no 7 in R9C3
2. Right-hand total in R7 = 20 must contain at least three digits with the lowest a maximum of 5 -> no 6,7 in R7C9, no 9 in R7C8
2a. Lower total in C9 = 30 contains 8,9 -> the remaining visible digits must total 13 = {256/346} (cannot be {1237/1246/1345/67} because of NC, cannot be {157/247} because 7 only in R89C9) = [6.52/64.3] -> R6C9 = 6, placed for C9, R9C9 = {23}, no 5 in R7C9, no 2,3,4,7 in R8C9
2b. 7 in C9 only in R23C9 -> no 6 in R2C8, no 6,7,8 in R3C8
Clean-ups:
R6C9 = 6 -> no 5,6 in R5C8, no 5 in R6C8, no 5,6,7 in R7C8
R9C9 = {23} -> no 2,3 in R89C8
3. Lower total in C8 = 24 must contain 9 (cannot be {24567/123567} because 7 only in R89C8, also can only be {789} with intermediate lower numbers) -> R3C8 = 9, placed for R3, no 8,9 in R8C8
3a. 7,8 of {789} must be in R79C8 -> no 7 in R8C8
3b. Total of 24 can only start with 1, 2, 3, 4 or 7 -> R9C8 = {147} (2,3 have been eliminated by clean-ups)
3c. Total of 24 cannot be {4569} because 5 only in R48C8 -> no 4 in R9C8
3d. Remaining possible ways to make total of 24 = {1689/789} (cannot be {12579/12678/13479} because 7 only in R9C8, cannot be {13569} because 5 only in R48C8, cannot be {12489} because 8 only in R8C8, cannot be {123459} because of NC) -> R7C8 = 8, placed for R7
3e. Right-hand total = 20 must be [9.83] -> R7C9 = 3, R89C9 = [52] (step 2a), 2,3,5 placed for C9, 3 placed for R7, 2 placed for R9, no 1 in R9C8
3f. R9C8 = 7, placed for R9
3g. R3C1 = 8, placed for R3 and C1, R6C1 = 9
3h. Upper total in C8 = 12 contains 9 in R3C8 = [319/339] (cannot be [129], NC) -> R1C8 = 3, placed for R1, R2C8 = {13}
Clean-ups:
R1C8 = 3 -> no 2,3,4 in R2C7, no 4 in R2C9
R2C8 = {13} -> no 2 in R3C7
R3C1 = 8 -> no 7 in R24C12 + R3C2
R3C8 = 9 -> no 9 in R4C7
R6C1 = 9 -> no 8 in R6C2
R7C8 = 8 -> no 7,9 in R678C7
R7C9 = 3 -> no 2,3,4 in R6C8, no 4 in R8C8
R8C9 = 5 -> no 6 in R8C8
R9C9 = 2 -> no 1 in R8C8
R8C8 = 5 -> no 4,5,6 in R79C7, no 4,6 in R8C7
R7C7 = {12} -> no 1,2 in R68C67 + R7C6, no 1 in R6C8
R6C8 = 6 -> no 5,6,7 in R5C7, no 5 in R6C7
R8C7 = {35} -> no 4 in R789C6
4 in C9 only in R34C9 -> no 3,5 in R4C8
4. Right-hand total in C8 = 14, R8C9 = 5 -> can only be [95] including repeated or lower numbers
4a. Left-hand total in C8 = 11 contains 9 -> R8C1 = 2, placed for C1
4b. Left-hand total = [29], right-hand total = [95], possibly including repeated numbers -> R8C2 = {29} (left-hand total cannot contain 1 because of NC)
4c. 6,7,8 only possible in R8 between two 9s -> no 6,7,8 in R8C6, no 8 in R8C5
[This was where I went wrong on my earlier attempts, I’d mistakenly thought that 6,7,8 could be eliminated from all cells in R8.]
Clean-up:
R8C1 = 2 -> no 1 in R7C1, no 1,2 in R7C2, no 1,3 in R9C12
5. Left-hand total in R4 = 10 must include at least two numbers with the first not higher than 4 -> R4C1 = {134}
6. Left-hand total in R5 = 25 must use at least four cells and contain 9 (but not 8 which is in R5C9) with first digit not higher than 4 -> R5C1 = {134}, no 7 in R5C2, no 9 in R5C3
6a. R45C1 only contain 1,3,4, cannot be {34} (NC) -> 1 in R45C1, locked for C1
Clean-ups:
1 in R45C1 -> no 2 in R4C2, no 1,2 in R5C2
R45C1 contain one of 3,4 -> no 3,4 in R5C2
6b. Only remaining possible way to make 25 with R5C2 = {56} and containing 9 = {3679} (cannot be {4579} NC) -> R5C1 = 3, R5C2 = 6, both placed for R5, 3 also placed for C1
Clean-ups:
R5C1 = 3 -> no 4 in R4C1, no 3,4 in R4C2, no 2,3,4 in R6C2
R5C2 = 6 -> no 5 in R4C2, no 5,6,7 in R46C3, no 5,7 in R5C3 + R6C2
R4C1 = 1 -> no 1,2 in R3C2
7a. Left-hand total in R4 = 10 cannot start [16] -> R4C2 = 1, no 1,2 in R5C3
7b. Total = 10 cannot start [118] -> no 8 in R4C3
7c. R5C3 = 4, placed for R5 and C3
Clean-ups:
R4C2 = 1 -> no 1,2 in R3C3, no 2 in R4C3
R5C3 = 4 -> no 3 in R46C3, no 3,4,5 in R46C4, no 5 in R5C4
R5C8 = {12} -> no 1,2 in R45C7, no 1 in R4C9
8a. R4C9 = 4, placed for C1
8b. R5C7 = 9, placed for R5 and C7
8c. R7C7 = 2 (hidden single in C7), placed for R7, no 3 in R8C7
8d. R8C7 = 5, placed for C7
8e. R9C24 = {89} (hidden pair in R9)
8f. Right-hand total in R9 = 18 contains 2,7 = {279} -> R9C24 = [89]
Clean-ups:
R5C7 = 9 -> no 8,9 in R46C6
R7C7 = 2 -> no 3 in R6C67 + R8C6
R9C2 = 8 -> no 9 in R8C2
R9C4 = 9 -> no 8 in R8C4, no 9 in R8C5
R8C2 = 2 -> no 1 in R7C3, no 1,3 in R89C3
R8C7 = 5 -> no 5,6 in R79C6
R9C3 = {56} -> no 5 in R8C34
9a. Left-hand total in R8 = 11 = [29] (step 4b) -> R8C3 = 2, placed for C3
9b. Naked pair {13} in R9C67, locked for R9
9c. Left-hand total in R8 = 11 = [29] -> R8C4 = {29}
Clean-ups:
R8C3 = 2 -> no 1 in R7C4
R6C7 = {46} -> no 5 in R56C6
R7C6 = {79} -> no 8 in R6C5
R9C5 = {456} -> no 5 in R8C5
R9C6 = {13} -> no 2 in R8C5
10a. Left-hand total 22 in R7 must contain 9 but 8 is placed in R7C8 so only possible way is {679} (cannot be {1579} because 1 in R7 only in R7C5) with possible hidden lower numbers -> R7C1 = 6, placed for R7 and C1, no 5,7 in R7C2
10b. R7C2 = 4, placed for R7
10c. R1C1 = 7 (hidden single in C1), placed for R1
10d. R7C5 = 1 (hidden single in R7), placed for C5
Clean-ups:
R1C1 = 7 -> no 6 in R12C2
R7C1 = 6 -> no 6 in R6C2
R7C2 = 4 -> no 5 in R7C3
R7C5 = 1 -> no 1,2 in R6C4, no 2 in R6C5 + R8C4
R8C4 = 9 -> no 9 in R7C3
R7C3 = 7 -> no 8 in R6C3, no 6,7,8 in R6C4
R2C1 = {45} -> no 4,5 in R13C2
R1C2 = {12} -> no 1 in R12C3
R1C3 = {56} -> no 5,6 in R1C4 + R2C34, no 5 in R2C2
R9C5 = {456} -> no 5 in R8C6
11a. R7C346 = [759], 7 placed for C3
11b. R2C3 = 8 (hidden single in C3)
Clean-ups:
R2C3 = 8 -> no 7,9 in R2C4, no 7 in R3C4
R7C4 = 5 -> no 4,5,6 in R6C5, no 4,6 in R8C5
R7C6 = 9 -> no 9 in R6C5
Naked pair {19} in R46C3 -> no 1,2 in R5C4
12a. R5C4 = 7, placed for R5, no 7 in R6C5
12b. R68C5 = [37], placed for C5
12c. R5C5 = 5 (hidden single in R5), placed for C5
Clean-ups:
R5C4 = 7 -> no 6,8 in R4C45
R5C5 = 5 -> no 4 in R4C5, no 4,5,6 in R4C6, no 4,6 in R6C6
R6C5 = 3 -> no 2 in R5C6
R6C6 = 7 -> no 6 in R6C7
R5C6 = 1 -> no 2 in R4C5
13a. R6C7 = 4, placed for C7
13b. R2C5 = 8 (hidden single in C5)
13c. 7 in C7 only in R34C7 -> no 6 in R34C7, NC
13d. R2C7 = 6 (hidden single in C7), no 7 in R3C7
13e. R4C7 = 7 (hidden single in C7)
Clean-ups:
R2C7 = 6 -> no 5,6 in R1C6, no 5 in R2C6, no 5,6,7 in R3C6
R4C7 = 7 -> no 6 in R4C8
R3C7 = {13} -> no 2 in R234C6 + R4C8
2 in C5 only in R13C5 -> no 1,2,3 in R2C4, no 1,3 in R2C6
[I ought to have spotted this earlier.]
14a. R2C1 + R3C2 = [46] (cannot be [45/55], AK, FNC), 4 placed for C1, 6 placed for R3
14b. R9C135 = [564], 6 placed for C3, 4 placed for C5
14c. R1C3 = 5, placed for C3, R3C3 = 3, placed for R3
14d. R3C45679 = [52417], 2 placed for C5, 1 placed for C7, 7 placed for C9
Clean-ups:
R1C3 = 5 -> no 4 in R1C4 + R2C24
R1C5 = 6 -> no 6 in R2C6
R3C3 = 3 -> no 2,3 in R2C2, no 2 in R4C4
R3C5 = 2 -> no 1 in R4C4, no 1,3 in R4C6
R3C7 = 1 -> no 1 in R24C8
R2C2 = 1 -> no 2 in R1C2
15a. R1C246 = [124]
15b. R5C68 = [12]
15c. R9C67 = [13]
[Now to apply five three-in-a-rows and one four-in-a-row]
16a. Four-in-a-row only possible in R6C1234 = [9999], 9 placed for C3
16b. Left-hand total in R4 = 10 = [19] -> R4C4 = 9
This leaves five three-in-a-rows, 8s in R2, 1s in R4, 2s and 5s in R8 and 4s in C6