咱们想作一个输入一个n,咱们就创建n子棋,条件是n>=3,创建的棋盘为(长宽都为为n-2)*n的棋盘。下棋时,X先下,O后下,知道一方取胜,或者为和棋。java
package cn.ztl.eightEndOfChapter; import java.util.Scanner; public class newPrint { static boolean erro_xy = true; private static Scanner input; public static void main(String[] args) { int n = nPrint(); play(n); } //输入函数 private static int nPrint() { input = new Scanner(System.in); System.out.print("请输入想玩几子棋:"); int n = input.nextInt(); return n; } //游戏函数 private static void play(int n) { char[][] _chessBoard = new char[(n - 2) * n][(n - 2) * n]; CheckerBoard(n, _chessBoard); } private static void CheckerBoard(int n, char[][] qi) { // 棋盘 // 第一次打印棋盘 ,没有棋子 for (int i = 0; i < qi.length; i++) { for (int j = 0; j < qi.length; j++) { System.out.print("-----"); } System.out.println(); for (int j = 0; j < qi.length; j++) { System.out.print("|"); System.out.print(" " + qi[i][j] + " "); } System.out.println("|"); } for (int j = 0; j < qi.length; j++) { System.out.print("-----"); } System.out.println(); boolean times = true;//设置计数 看是白棋仍是黑棋 while (true) {//屡次循环 每次为游戏者输入 char[][] newQi; newQi = chessBoard(chessPlayer((n - 2) * n, times), qi, n, times);// 得到新棋谱 if (newQi == null)//若是为null 那么结束游戏 break; if (!erro_xy) {// 下棋处已经存在棋子,报错 times = !times; erro_xy = true; } times = !times;//换棋子 qi = newQi;// 转换身份 } } private static int judgeToWin(int[] a, char[][] qi, int n, boolean times) {//判断玩家是否已经取胜或为和棋 char temp = (times) ? 'X' : 'O'; int DX[] = { 1, 0, 1, 1 }; int DY[] = { 0, 1, 1, -1 }; int c1, c2, xx, yy; for (int k = 0; k < 4; k++) { c1 = c2 = 0; for (xx = a[0] + DX[k], yy = a[1] + DY[k]; xx >= 0 && yy >= 0 && xx < qi.length && yy < qi.length && qi[xx][yy] == temp; xx += DX[k], yy += DY[k]) { c1++; } for (xx = a[0] - DX[k], yy = a[1] - DY[k]; xx >= 0 && yy >= 0 && xx < qi.length && yy < qi.length && qi[xx][yy] == temp; xx -= DX[k], yy -= DY[k]) { c2++; } if (c1 + c2 >= n - 1) return (times) ? 1 : 2; } return heqi(qi); } private static char[][] chessBoard(int[] a, char[][] newQi, int n, boolean times) {//将下棋人下的棋存入棋谱,并画新棋谱 if (newQi[a[0]][a[1]] == '\u0000') newQi[a[0]][a[1]] = (times) ? 'X' : 'O'; else { System.out.println("已经有棋子了,请从新输入"); erro_xy = false; } for (int i = 0; i < newQi.length; i++) { for (int j = 0; j < newQi.length; j++) { System.out.print("----"); } System.out.println(); for (int j = 0; j < newQi.length; j++) { System.out.print("|"); System.out.print(" " + newQi[i][j] + " "); } System.out.println("|"); } for (int j = 0; j < newQi.length; j++) { System.out.print("----"); } System.out.println(); int key = judgeToWin(a, newQi, n, times); if (key == 1) { System.out.println("A棋手赢了"); return null; } else if (key == 2) { System.out.println("B棋手赢了"); return null; } else if (key == 3) { System.out.println("和棋"); return null; } else { System.out.println(); } return newQi; } private static int heqi(char[][] qi) {//和棋判断 int sum = 0; for (int i = 0; i < qi.length; i++) { for (int j = 0; j < qi[i].length; j++) { if (qi[i][j] == '\u0000') sum++; } } if (sum == 0) return 3; else return 0; } private static int[] chessPlayer(int n, boolean times) {//下棋者 输入函数 input = new Scanner(System.in); int[] a = new int[2]; char temp = (times) ? 'A' : 'B'; char temp2 = (times) ? 'X' : 'O'; System.out.print("请棋手" + temp + "(" + temp2 + ")先下棋:x轴范围1-" + n + " :"); a[0] = input.nextInt() - 1; System.out.print("请棋手" + temp + "(" + temp2 + ")先下棋:y轴范围1-" + n + " :"); a[1] = input.nextInt() - 1; return a; } }
1.代码写的比较臃肿,由于还有事作,就先不简化了,有时间再作简化和图形化界面。
2.问题,不少东西能够写在方法外面的,一开始没想清楚就开写,致使最后不想改了。
3.我认为三目运算符很好用,再处理黑白棋时,如char temp2 = (times) ? ‘X’ : ‘O’;无论是打印时,仍是在比较时,用它能够将黑棋白棋认为是一种棋。
4.用boolean 类型的,每次取反,能够表现两种相反的状态。
5.在计算多子同线问题时,要善用for循环(开始时的条件;循环成立的条件;每次循环的变化)web
咱们接收到下棋者输入的,xy坐标,放入坐标系中原本是要考虑8个方向,咱们先考虑四个方向,固然还要考虑数组越界的状况,这四个方向都是xy坐标经过加0,加1,加-1能够到达的方向数组
for (xx = a[0] + DX[k], yy = a[1] + DY[k]; xx >= 0 && yy >= 0 && xx < qi.length && yy < qi.length && qi[xx][yy] == temp; xx += DX[k], yy += DY[k]) { c1++; }
然后面那个,考虑的是相反的四个方向恰好考虑彻底。若是一条线上超过n-1个那就说明已经赢了。svg