Java 小游戏 - 井字棋 v1.0 (初步完成) …

2019-12-15 16:04:40来源:博客园 阅读 ()

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Java 小游戏 - 井字棋 v1.0 (初步完成) (2018.4.16更新)

 
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; //
  9     static int rank; //
 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 }

 


原文链接:https://www.cnblogs.com/wangjiaolong/p/12044484.html
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