我正在做一个TicTacToe计划，唯一缺少的部分是让用户选择是退出游戏还是重播。我无法找到回归游戏的方法。
import java.io.*;
public class Expierment
{
static char c1 [] = new char[10];
static char c2 [] = new char[10];
static char c3 [] = new char[10];
static char p1;
static char p2;
static boolean gameOver = false;
public static void main(String args[])
{
int counter = 0;
int p1Wins = 0;
int p2Wins = 0;
int r1 = 0;
int r2 = 0;
int r3 = 0;
int r4 = 0;
int r5 = 0;
int r6 = 0;
int r7 = 0;
int r8 = 0;
int r9 = 0;
int pick1 = 0;
int pick2 = 0;
int pick3 = 0;
int pick4 = 0;
int pick5 = 0;
int pick6 = 0;
int pick7 = 0;
int pick8 = 0;
int pick9 = 0;
char turn = 'X';
int choice = menu();
switch(choice)
{
case 1:
System.out.println("The game is called 'Tic-Tac-Toe', you should have known it. If you don't, search it.") ;
case 2:
gameOver = false;
break;
case 3:
System.out.println("\nSee you next time !!");
return;
default:
System.out.println("\nYou hit the wrong key......\n");
return;
}//end of switch
System.out.println("\nPlayer 1 initials ?");
String n1 = GCS();
p1 = n1.charAt(0);
System.out.println("\nPlayer 2 initials ?");
String n2 = GCS();
p2 = n2.charAt(0);
c1[2]='1';
c2[2]='2';
c3[2]='3';
c1[1]='4';
c2[1]='5';
c3[1]='6';
c1[0]='7';
c2[0]='8';
c3[0]='9';
printBoard ();
while(gameOver!=true)
{
System.out.println("Which spot ?");
int pick = Integer. parseInt(GCS());
switch (pick)
{
case 1:
if (r1<1)
{
c1[2] = turn;
r1++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 2:
if (r2<1)
{
c2[2] = turn;
r2++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 3:
if (r3<1)
{
c3[2] = turn;
r3++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 4:
if (r4<1)
{
c1[1] = turn;
r4++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 5:
if (r5<1)
{
c2[1] = turn;
r5++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 6:
if (r6<1)
{
c3[1] = turn;
r6++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 7:
if (r7<1)
{
c1[0] = turn;
r7++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 8:
if (r8<1)
{
c2[0] = turn;
r8++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
case 9:
if (r9<1)
{
c3[0] = turn;
r9++;
}
else
{
System.out.println("That column is full, pick another.\n");
continue;
}
break;
default:
System.out.println("Seriously?! Pick a possible spot.\n");
continue;
}//end of switch
if (turn=='X') turn = 'O';
else turn = 'X';
printBoard();
if (checkWinner())
{
while(gameOver==true)
{
int Echoice = EGM();
switch(Echoice)
{
case 1:
System.out.println("The game is called 'Tic-Tac-Toe', you should have known it. If you don't, search it.") ;
case 2:
gameOver = false;
menu();
break;
case 3:
System.out.println("\nSee you next time !!");
return;
default:
System.out.println("\nYou hit the wrong key......\n");
return;
}//end of switch
}//end of while true
return;
}
counter ++;
if (counter==9)
{
System.out.println("\n\nYou tied.\n");
return;
}
}//end of while not true
}//end of main
public static boolean checkWinner()
{
for (int k=0; k<2; k++)
{
if ((c1[k]!=' ')&&(c1[k]==c2[k])&&(c1[k]==c3[k]))
{
System.out.println("\nYo " + c1[k] + " is the winner!\n");
gameOver=true;
return true;
}//checks column 1-3 horizontally
}//end of horizontal check
for (int m=0; m<2; m++)
{
if((c1[m]!=' ')&&(c1[m]==c1[m+1])&&(c1[m+1]==c1[m+2])&&(c1[m]==c1[m+2]))
{
System.out.println("\nYo " + c1[m] + " is the winner!\n");
gameOver=true;
return true;
}//checks column 1 vertically
if((c2[m]!=' ')&&(c2[m]==c2[m+1])&&(c2[m+1]==c2[m+2])&&(c2[m]==c2[m+2]))
{
System.out.println("\nYo " + c2[m] + " is the winner!\n");
gameOver=true;
return true;
}//checks column 2 vertically
if((c3[m]!=' ')&&(c3[m]==c3[m+1])&&(c3[m+1]==c3[m+2])&&(c3[m]==c1[m+2]))
{
System.out.println("\nYo " + c3[m] + " is the winner!\n");
gameOver=true;
return true;
}//checks column 3 vertically
if ((c1[m]!=' ')&&(c1[m]==c2[m+1])&&(c1[m]==c3[m+2]))
{
System.out.println("\nYo " + c1[m] + " is the winner!\n");
gameOver=true;
return true;
}//checks upward diagonal
if ((c3[m]!=' ')&&(c3[m]==c2[m+1])&&(c3[m]==c1[m+2]))
{
System.out.println("\nYo " + c1[m] + " is the winner!\n");
gameOver=true;
return true;
}
}//end of vertical check
return false;
}//end of checkWinner
public static void printBoard()
{
System.out.println("_______");
for (int j = 2; j > -1; j--)
{
System.out.println("|" + c1[j] + "|" + c2[j] + "|" + c3[j] + "|");
System.out.println("-------");
}
}//end of printBoard
public static int menu()
{
System.out.println("Tic-Tac-Toe ~ Main Menu\n\n1. Instructions\n2. Play a 1 player game"+"\n3. Exit\n");
int selection = Integer.parseInt(GCS());
return selection;
}//end of menu
public static int EGM()
{
System.out.println("Tic-Tac-Toe ~ End of Game Menu\n\n1. Instructions\n2. Play again"+"\n3. Exit\n");
int Eselection = Integer.parseInt(GCS());
return Eselection;
}
public static String GCS()
{
int noMoreInput=-1;
char enterKeyHit='\n';
int InputChar;
StringBuffer InputBuffer = new StringBuffer(100);
try
{
InputChar=System.in.read();
while(InputChar != noMoreInput)
{
if((char)InputChar !=enterKeyHit)
{
InputBuffer.append((char)InputChar);
}
else
{
InputBuffer.setLength(InputBuffer.length()-1);
break;
}
InputChar=System.in.read();
}
}
catch (IOException IOX)
{
System.err.println(IOX);
}
return InputBuffer.toString();
}//end of GCS
}//end of public class
答案 0 :(得分：2)
你真的应该从main函数中获取一些代码。
具体来说，我将整个游戏循环放在一个单独的函数中，可能称为playGame()，其中包含玩游戏，检查获胜者等的逻辑，并返回获胜者或只是打印获胜者并返回无效。
然后main函数可以在循环中调用playGame()，并在循环结束时询问用户他/她是否想再次播放。
通常，您希望每个功能都执行一个逻辑任务。移动checkWinner已经做得很好，现在对其他一些代码执行相同的操作。
如果您在“要求用户再次播放”方面需要帮助，请发表评论，然后我会进行编辑以解决此问题。
答案 1 :(得分：1)
quick and dirty pseudo-code - not modular
do {
//everything in your main goes here
.
.
.
playAgain = prompt the user
} while(playAgain);
答案 2 :(得分：1)
比do-while方法更脏。最终会导致堆栈溢出。
//your current main method
boolean playAgain = prompt the user
if(playAgain){
main(args);
}
答案 3 :(得分：1)
使用当前的程序布局，没有“干净”的方法来实现它。以下是一些建设性的批评：
您应该只使用main方法初始化您的程序。因此，尝试 1 方法仅执行 1 事情。你现在设计的游戏菜单和游戏主循环都在同一个方法中。
您的游戏循环可能如下所示：
while still playing
read input from user
if game is active
process game input
update game
else
process menu input
update menu
这样，你只需要为菜单或游戏开启game is active状态。等
您的GCS方法过于复杂，只需将其替换为：
Scanner scanner = new Scanner(); // put this somewhere at the class level (so it is reusable)
...
String input = scanner.nextLine(); // put this somewhere in a method reading an input
可能（可能像其他人建议的那样）使用数组，或者更具体地说是二维数组，而不是初始化许多变量。
int grid[][] = new int[3][3];
// grid[0][0] points to the top-left cell, grid[2][2] points to the bottom right one
或者您可以使用单个int来存储所有内容;将您的网格表示为bit array
int grid = 0; // empty grid
...
// set player move
grid |= (1 << (y*3)+x) << PLAYER_OFFSET; // PLAYER_OFFSET: 0=player 1, 16=player 2
// reset (clear) player move
grid &= ~((1 << (y*3)+x) << PLAYER_OFFSET); // ...
// check if player move is set
boolean isSet = (grid >> ((1 << (y*3)+x) << PLAYER_OFFSET)) && 1; // ...
为什么位阵列很酷？因为要检查玩家是否获胜，你不需要花哨的for..loop和其他复杂的算法，只需要对获胜模式进行验证......：
int pattern = 273; // test for diagonal from [0,0] to [2,2], or bits "100 010 001"
boolean isWinning = (grid && (pattern << PLAYER_OFFSET)) != 0;
就是这样！
此外，使用0到9可能很适合识别网格单元格，但是因为输入值不是 直观的玩家。例如，许多游戏（象棋，西洋跳棋，LOA等）和应用程序（Excel，Calc等）都在使用Algebraic chess notation系统。转换x和y中的表示法非常简单。例如：
boolean cellPlayed = false;
while (!cellPlayed) {
String cellStr = scanner.readLine().toLower(); // ex: "b2" for the center cell
try {
int gridx = cellStr.charAt(0) - 'a'; // ex: for "b2" should return 1
int gridy = cellStr.charAt(1) - '1'; // ex: for "b2" should return 1
grid(gridx][gridy] = playerValue; // 1 for "player 1" and 2 for "player 2"
cellPlayed = true;
} catch (Exception e) {
System.out.println("Error! Invalid input");
}
}
不要气馁！我们都从某个地方开始！ ：）
答案 4 :(得分：0)
使用更简洁的主要方法可以更加清晰。真正快速和肮脏的方法是将main方法中的所有内容复制到一个新方法中，然后从主要方法调用该方法，并在用户选择开始新游戏时使用。