井字棋游戏初步完成
实现功能：输入位置数据->打印棋盘->判断是否胜利->继续游戏/退出游戏
缺点：没有清屏函数 判断胜利方法太过无脑
1 package MYGAME;
2
3 import java.util.Scanner;
4
5 public class Mygame {
6 static int winnerx = 0;
7 static int winnero = 0;
8 static int row; // 行
9 static int rank; // 行
10 static int[][] x = new int[3][3]; // X玩家的棋子
11 static int[][] o = new int[3][3]; // O玩家的棋子
12
13 public static void main(String[] args) {
14 inputmap();
15 for (int p = 0; p < 9; p++) {
16 if (p % 2 == 1) {
17 oplay();
18 if (winnero==1){
19 System.out.println("o选手胜利");
20 break;
21 }
22 } else {
23 xplay();
24 if (winnerx==1){
25 System.out.println("x选手胜利");
26 break;
27 }
28 }
29 }
30 System.out.println("游戏结束");
31 }
32
33 // 输出棋盘
34 public static void inputmap() {
35 System.out.println(" ");
36 for (int m = 0; m < 3; m++) {
37 for (int n = 0; n < 3; n++) {
38 if (x[m][n] == 1) {
39 System.out.print("x|");
40 } else if (o[m][n] == 1) {
41 System.out.print("o|");
42 } else {
43 System.out.print(" |");
44 }
45 }
46 System.out.println(" ");
47 }
48 }
49
50 // x选手开始下棋
51 public static void xplay() {
52 Scanner input = new Scanner(System.in);
53 System.out.print("轮到x选手下棋\n");
54 System.out.print("请选择第几行\n");
55 row = input.nextInt() - 1;
56 System.out.print("请选择第几列\n");
57 rank = input.nextInt() - 1;
58 x[row][rank] = 1;
59 inputmap();
60 win();
61 }
62
63 // o选手开始下棋
64 public static void oplay() {
65 Scanner input = new Scanner(System.in);
66 System.out.print("轮到o选手下棋\n");
67 System.out.print("请选择第几行\n");
68 row = input.nextInt() - 1;
69 System.out.print("请选择第几列\n");
70 rank = input.nextInt() - 1;
71 o[row][rank] = 1;
72 inputmap();
73 win();
74 }
75
76 // 手动清屏=。=
77 public static void cleanscreen() {
78 for (int i = 0; i < 50; i++) {
79 System.out.println("");
80 }
81 }
82
83
84 //判断是否胜利
85 public static void win() {
86 if (
87 (x[0][0] == 1 & x[1][0] == 1 & x[2][0] == 1)||
88 (x[0][1] == 1 & x[1][1] == 1 & x[2][1] == 1)||
89 (x[0][2] == 1 & x[1][2] == 1 & x[2][2] == 1)||
90 (x[0][0] == 1 & x[0][1] == 1 & x[0][2] == 1)||
91 (x[1][0] == 1 & x[1][1] == 1 & x[1][2] == 1)||
92 (x[2][0] == 1 & x[2][1] == 1 & x[2][2] == 1)||
93 (x[0][0] == 1 & x[1][1] == 1 & x[2][2] == 1)||
94 (x[0][2] == 1 & x[1][1] == 1 & x[2][0] == 1)
95 ){
96 winnerx=1;
97 }
98 if (
99 (o[0][0] == 1 & o[1][0] == 1 & o[2][0] == 1)||
100 (o[0][1] == 1 & o[1][1] == 1 & o[2][1] == 1)||
101 (o[0][2] == 1 & o[1][2] == 1 & o[2][2] == 1)||
102 (o[0][0] == 1 & o[0][1] == 1 & o[0][2] == 1)||
103 (o[1][0] == 1 & o[1][1] == 1 & o[1][2] == 1)||
104 (o[2][0] == 1 & o[2][1] == 1 & o[2][2] == 1)||
105 (o[0][0] == 1 & o[1][1] == 1 & o[2][2] == 1)||
106 (o[0][2] == 1 & o[1][1] == 1 & o[2][0] == 1)
107 ){
108 winnero=1;
109 }
110 }
111
112 }
来源：oschina
链接：https://my.oschina.net/u/4266515/blog/3329985