我正在尝试做一个简单的游戏（X和O）问题，当其中一个玩家赢了游戏时，另一个玩家仍然可以玩...当他们中的一个赢得游戏时我怎么能结束游戏呢？ / p>
fun buClicked(view: View){
var buChoise = view as Button
var cellID =0
when (buChoise.id){
R.id.b1-> cellID=1
R.id.b2-> cellID=2
R.id.b3-> cellID=3
R.id.b4-> cellID=4
R.id.b5-> cellID=5
R.id.b6-> cellID=6
R.id.b7-> cellID=7
R.id.b8-> cellID=8
R.id.b9-> cellID=9
}
PlayGame(cellID,buChoise)
}
var Player1=ArrayList<Int>()
var Player2=ArrayList<Int>()
var activePlayer=1
fun PlayGame(cellID:Int,buChoise:Button){
if(activePlayer==1){
buChoise.text="X"
buChoise.setBackgroundResource(R.color.colorPrimaryDark)
Player1.add(cellID)
activePlayer=2
}else{
buChoise.text="O"
buChoise.setBackgroundResource(R.color.colorAccent)
Player2.add(cellID)
activePlayer=1
}
buChoise.isEnabled=false
checkWinner()
}
fun checkWinner(){
var winner= -1
//row 1
if(Player1.contains(1) && Player1.contains(2) && Player1.contains(3)){
Toast.makeText(this,"the winner is player 1",Toast.LENGTH_LONG).show()
}
if(Player2.contains(1) && Player2.contains(2) && Player2.contains(3)){
Toast.makeText(this,"the winner is player 2",Toast.LENGTH_LONG).show()
}
//row 2
if(Player1.contains(4) && Player1.contains(5) && Player1.contains(6)){
Toast.makeText(this,"the winner is player 1",Toast.LENGTH_LONG).show()
}
if(Player2.contains(4) && Player2.contains(5) && Player2.contains(6)){
Toast.makeText(this,"the winner is player 2",Toast.LENGTH_LONG).show()
}
//row 3
if(Player1.contains(7) && Player1.contains(8) && Player1.contains(9)){
Toast.makeText(this,"the winner is player 1",Toast.LENGTH_LONG).show()
}
if(Player2.contains(7) && Player2.contains(8) && Player2.contains(9)){
Toast.makeText(this,"the winner is player 2",Toast.LENGTH_LONG).show()
}
//col 1
if(Player1.contains(1) && Player1.contains(4) && Player1.contains(7)){
Toast.makeText(this,"the winner is player 1",Toast.LENGTH_LONG).show()
}
if(Player2.contains(1) && Player2.contains(4) && Player2.contains(7)){
Toast.makeText(this,"the winner is player 2",Toast.LENGTH_LONG).show()
}
//row 2
if(Player1.contains(2) && Player1.contains(5) && Player1.contains(8)){
Toast.makeText(this,"the winner is player 1",Toast.LENGTH_LONG).show()
}
if(Player2.contains(2) && Player2.contains(5) && Player2.contains(8)){
Toast.makeText(this,"the winner is player 2",Toast.LENGTH_LONG).show()
}
//col 3
if(Player1.contains(3) && Player1.contains(6) && Player1.contains(9)){
Toast.makeText(this,"the winner is player 1",Toast.LENGTH_LONG).show()
}
if(Player2.contains(3) && Player2.contains(6) && Player2.contains(9)){
Toast.makeText(this,"the winner is player 2",Toast.LENGTH_LONG).show()
}
}
答案 0 :(得分：0)
if (Player1.containsAll(listOf(1, 2, 3)) || Player1.containsAll(listOf(4, 5, 6)) || Player1.containsAll(listOf(7, 8, 9)) || Player1.containsAll(listOf(1, 5, 9)) || Player1.containsAll(listOf(3, 5, 7))) {
Toast.makeText(this, "the winner is player 1", Toast.LENGTH_LONG).show()
} else if (Player2.containsAll(listOf(1, 2, 3)) || Player2.containsAll(listOf(4, 5, 6)) || Player2.containsAll(listOf(7, 8, 9)) || Player2.containsAll(listOf(1, 5, 9)) || Player2.containsAll(listOf(3, 5, 7))) {
Toast.makeText(this, "the winner is player 2", Toast.LENGTH_LONG).show()
}
答案 1 :(得分：0)
“当其中一个玩家赢了游戏时，另一个玩家仍然可以玩”的问题。你只需要一个布尔变量。在类的顶部添加var isGameOver。
fun main(args: Array<String>) {
var isGameOver = false
fun checkWinner(){
var winner= -1
//row 1
if(Player1.contains(1) && Player1.contains(2) && Player1.contains(3)){
isGameOver = true
}
..........
}
fun PlayGame(cellID:Int,buChoise:Button){
if(!isGameOver){
if(activePlayer==1){
buChoise.text="X"
buChoise.setBackgroundResource(R.color.colorPrimaryDark)
Player1.add(cellID)
activePlayer=2
}else{
buChoise.text="O"
buChoise.setBackgroundResource(R.color.colorAccent)
Player2.add(cellID)
activePlayer=1
}
buChoise.isEnabled=false
checkWinner()
}else{
//give replay button
}
}
}
然后在PlayGame功能上给出条件。如果isGameOver为true，则两个玩家都无法再次执行操作并给他们重播按钮以重复此过程。