Input:
[
["8","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being
modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
Note:
A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.
The given board contain only digits 1-9 and the character '.'.
The given board size is always 9x9.
Analyse:
Solution:
classSolution {funcisValidSudoku(_board: [[Character]]) ->Bool {for i in0..<board.count {let isValid = self.checkIsSudoku(i1: i, i2: i +1, j1:0, j2:board[i].count, board);if(isValid ==false) {returnfalse; } }for i in0..<board[0].count {let isValid = self.checkIsSudoku(i1:0, i2: board.count, j1:i, j2:i +1, board);if(isValid ==false) {returnfalse; } }for i in0..<3 {for j in0..<3 {let isValid = self.checkIsSudoku(i1: i *3, i2: i *3+3, j1: j *3, j2: j *3+3, board);if(isValid ==false) {returnfalse; } } }returntrue; }funccheckIsSudoku(i1: Int, i2: Int, j1:Int, j2:Int, _board: [[Character]]) ->Bool {var array:[Character] = [];for i in i1..<i2 {for j in j1..<j2 {let value = board[i][j];if(value ==".") {continue; }if(array.index(of: value)!=nil) {// print("\(value), \(array.index(of: value))");returnfalse; }else { array.append(value); } } }returntrue; }}