题解 | #井字棋#
井字棋
https://www.nowcoder.com/practice/e1bb714eb9924188a0d5a6df2216a3d1
import java.util.*; public class serve { public static boolean checkWon(int[][] board) { int[] chessman = new int[9]; int s = 0; for (int j = 0; j <= 2; j++) { for (int i = 0; i <= 2; i++) { chessman[s] = board[j][i]; s++; } } if (chessman[0] == chessman[1] && chessman[1] == chessman[2] && chessman[0] == 1) { return true; } else if (chessman[3] == chessman[4] && chessman[4] == chessman[5] && chessman[3] == 1) { return true; } else if (chessman[6] == chessman[7] && chessman[7] == chessman[8] && chessman[6] == 1) { return true; } else if (chessman[0] == chessman[3] && chessman[3] == chessman[5] && chessman[0] == 1) { return true; } else if (chessman[1] == chessman[4] && chessman[4] == chessman[7] && chessman[1] == 1) { return true; } else if (chessman[2] == chessman[5] && chessman[8] == chessman[5] && chessman[2] == 1) { return true; } else if (chessman[0] == chessman[4] && chessman[4] == chessman[8] && chessman[0] == 1) { return true; } else if (chessman[2] == chessman[4] && chessman[4] == chessman[6] && chessman[2] == 1) { return true; } else { return false; } } public static void main(String[] args) { int[][] board = {{-1,1,0},{1,-1,1},{0,1,-1}}; boolean result = checkWon(board); System.out.println(result); } }
这个是我自己的想法,我选择先将二维数组中的代码保存至一个一维数组中只需判断,一维数组中相邻三个元素以及对角线上的元素是否都为一,如果都为一返回true,反之返回false;
import java.util.*; public class Board { public static boolean checkWon(int[][] board) { // 检查行 for (int i = 0; i < 3; i++) { if (board[i][0] == 1 && board[i][1] == 1 && board[i][2] == 1) { return true; } } // 检查列 for (int j = 0; j < 3; j++) { if (board[0][j] == 1 && board[1][j] == 1 && board[2][j] == 1) { return true; } } // 检查对角线 if ((board[0][0] == 1 && board[1][1] == 1 && board[2][2] == 1) || (board[0][2] == 1 && board[1][1] == 1 && board[2][0] == 1)) { return true; } return false; } }
这个是直接对二维数组直接进行检察,显然这个方法更简单,内存使用更少,速度更快
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