Valid Sudoku in Swift
This is my solution to LeetCode – Valid Sudoku in Swift.
class Solution {
func isValidSudoku(_ board: [[Character]]) -> Bool {
let count = board.count
var set = Set<Character>()
for i in 0..<count{
// firstly
set = Set(board[i])
var num = board[i].reduce(0 , {(result : Int, char : Character)
in
var cent = 0
if String(char) == "."{
cent = 1
}
return result + cent
})
if num > 0 , count - num != set.count - 1{
return false
}
else if num == 0, set.count != count{
return false
}
// secondly
set = Set(board.reduce([Character]() , { resultArray, chars in
return resultArray + [chars[i]]
}))
num = board.reduce(0 , {(result : Int, chars : [Character])
in
var cent = 0
if String(chars[i]) == "."{
cent = 1
}
return result + cent
})
if num > 0 , count - num != set.count - 1{
return false
}
else if num == 0, set.count != count{
return false
}
// thirdly
let characters = board.flatMap{
return 0ドル
}
let fisrtMiddle = ( i/3 ) * 27 + ( i % 3 ) * 3 + 1
let secondMiddle = fisrtMiddle + 9
let thirdMiddle = fisrtMiddle + 18
let arrayThree = [characters[fisrtMiddle - 1], characters[fisrtMiddle], characters[fisrtMiddle + 1],
characters[secondMiddle - 1], characters[secondMiddle], characters[secondMiddle + 1],
characters[thirdMiddle - 1], characters[thirdMiddle], characters[thirdMiddle + 1]]
set = Set(arrayThree)
num = arrayThree.reduce(0 , {(result : Int, char : Character)
in
var cent = 0
if String(char) == "."{
cent = 1
}
return result + cent
})
if num > 0 , count - num != set.count - 1{
return false
}
else if num == 0, set.count != count{
return false
}
}
return true
}
}
How can I make it shorter in syntax and keep intuitive, like the following Python code?
def isValidSudoku(self, board):
seen = sum(([(c, i), (j, c), (i/3, j/3, c)]
for i, row in enumerate(board)
for j, c in enumerate(row)
if c != '.'), [])
return len(seen) == len(set(seen))
The Python code is very Pythonic and short.
How to use Swift syntax power to make my code shorter?
I think Functional is a good choice. RxSwift is welcomed.
dengApro
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