Question

I am making a tic tac toe game in Swift for 2 players. Now I know I can find a winner by making outlets for the 9 buttons and then writing for all the 3 in row cases, but I was wondering if there is a better way to do this? And is there a better way to write my switch statement?

ReDim Apps(size - 1)

For i = 1 To size
    Apps(i - 1) = Cells(i, 1).Value
Next i

Answer 1

我会在按钮上使用值0-9设置[ 0 ][ 1 ][ 2 ] [ 3 ][ 4 ][ 5 ] [ 6 ][ 7 ][ 8 ]属性，按从上到下，从左到右的顺序排列，如下所示：

viewForTag(_:)

然后创建一个模型来表示每个方块并给它一个相应的索引值，以及对拥有它的玩家的引用。然后，您可以使用这些模型处理游戏逻辑，并通过使用class Square { var owningPlayer: Player? let index: Int init( index: Int ) { self.index = index } } class Player { let symbol: String init( symbol: String ) { self.symbol = symbol } } class ViewController: UIViewController { var squares = [Square]() var players = [Player]() var currentPlayer: Player? override func viewDidLoad() { super.viewDidLoad() self.players = [ Player(symbol: "X"), Player(symbol: "O") ] self.currentPlayer = self.players[0] // Create squares for i in 0..<9 { self.squares.append( Square(index: i) ) } } @IBAction func buttonPressed( button: UIButton ) { guard let player = self.currentPlayer else { fatalError( "Don't have a current player!" ) } if let square = self.squares.filter({ $0.index == button.tag }).first { // Update model square.owningPlayer = player // Update UI button.setTitle( player.symbol, forState: .Normal ) } if self.checkWin() { print( "\(player.symbol) wins!" ) } else { self.currentPlayer = self.players.filter({ $0 !== player }).first } } func checkWin() -> Bool { guard let player = self.currentPlayer else { fatalError( "Don't have a current player!" ) } let winningSequences = [ // Horizontal rows [ 0, 1, 2 ], [ 3, 4, 5 ], [ 6, 7, 8 ], // Diagonals [ 0, 4, 8 ], [ 2, 4, 6 ], // Vertical rows [ 0, 3, 6 ], [ 1, 4, 7 ], [ 2, 5, 8 ], ] // Get indexes owned by this player let selectedIndexes = self.squares.filter({ $0.owningPlayer === player }).map { $0.index } // Change the sequence arrays into sets for accurate comparison using `contains(_:)` if winningSequences.map({ Set<Int>($0) }).contains( Set<Int>(selectedIndexes) ) { return true } return false } }获取对按钮的引用来更新UI。这是我的版本：

if(Boolean){
    if(Boolean) something();
    else if(Boolean) something(); 
    else something();
           }

Answer 2

虽然@Patrick Lynch的答案非常出色，但我发现如果比赛超过3次，它就无法与胜利相提并论。

这是因为他只检查任何winsSequences是否包含当前移动序列（selectedIndexes）。由于winsSequences地图最多只有3个项目，因此它永远不会包含一组4个动作。

要解决此问题，我将包含更改为子集。即winsSequence（0,1,2）selectedIndexes（0,5,1,2）的子集

替换

if winningSequences.map({ Set<Int>($0) }).contains( Set<Int>(selectedIndexes) ) { 
 return true 
}

与

for seq in winningSequences.map( { Set<Int>($0)}) {
  if seq.isSubset(of: selectedIndexes) {
     return true
     }
  }

我还是新手，所以如果我错了或任何人有更好的主意，那么我会很高兴听到它。

Tic tac toe finding win?

2 个答案: