力扣：794. 有效的井字游戏

what is an object

于 2022-07-13 22:39:18 发布

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分类专栏：算法菜鸟刷题记录文章标签： leetcode 游戏 java

本文链接：https://blog.youkuaiyun.com/newOneObject/article/details/125774286

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本文探讨了如何用Java实现Tic Tac Toe游戏的解决方案，涉及游戏板状态检查、X和O计数以及判断胜负条件。重点在于算法逻辑和游戏规则的实现细节。

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模拟题，不想写，CV一下，


/**
 * @author xnl
 * @Description:
 * @date: 2022/7/13   22:11
 */
public class Solution {
    public static void main(String[] args) {
        Solution solution = new Solution();

    }

    public boolean validTicTacToe(String[] board) {
        int xCount = 0 , yCount = 0;
        for (String row : board) {
            for (char c : row.toCharArray()) {
                xCount = (c == 'X') ? xCount + 1 : xCount;
                yCount = (c == 'O') ? yCount + 1 : yCount;
            }
        }
        return !((yCount != xCount && yCount != xCount - 1) ||
                (yCount != xCount - 1 && win(board, 'X')) ||
                (yCount != xCount && win(board, 'O')));
    }

    public boolean win(String[] board, char p){
        for (int i = 0; i < 3; i++){
            if (((p == board[0].charAt(i)) && (p == board[1].charAt(i)) && (p == board[2].charAt(i))) ||
                    (p == board[i].charAt(0) && p == board[i].charAt(1) && p == board[i].charAt(2))){
                return true;
            }
        }
        return (p == board[0].charAt(0) && p == board[1].charAt(1) && p == board[2].charAt(2)) ||
                (p == board[0].charAt(2) && p == board[1].charAt(1) && p == board[2].charAt(0));
    }
}