设计一个算法,判断玩家是否赢了井字游戏。输入是一个 N x N 的数组棋盘,由字符" ","X"和"O"组成,其中字符" "代表一个空位。
以下是井字游戏的规则:
示例 1 :
输入: board = ["O X"," XO","X O"] 输出: "X"
示例 2 :
输入: board = ["OOX","XXO","OXO"] 输出: "Draw" 解释: 没有玩家获胜且不存在空位
示例 3 :
输入: board = ["OOX","XXO","OX "] 输出: "Pending" 解释: 没有玩家获胜且仍存在空位
提示 :
1 <= board.length == board[i].length <= 100 输入一定遵循井字棋规则
使用求和的方法,获取到横、纵、对角线的值,判断对应的值是否达成了全部一致
class Solution { public String tictactoe(String[] board) { int length = board.length; int heng = 0; int zong = 0; int left = 0; int right = 0; //是否出现空格 boolean flag = false; for(int i = 0;i<length;i++){ heng = 0; zong = 0; for(int j = 0;j<length;j++){ //记录横纵的值 heng = heng + (int)board[i].charAt(j); zong = zong + (int)board[j].charAt(i); if(board[i].charAt(j) == ' '){ flag = true; } } //检查横纵 if(heng == (int)'X' * length || zong == (int)'X' * length){ return "X"; } if(heng == (int)'O' * length || zong == (int)'O' * length){ return "O"; } //记录斜线的值 left = left + board[i].charAt(i); right = right + board[i].charAt(length-i-1); } //检查对角线 if(left == (int)'X' * length || right == (int)'X' * length){ return "X"; } if(left == (int)'O' * length || right == (int)'O' * length){ return "O"; } if(flag){ return "Pending"; } return "Draw"; } }
原文链接:https://www.cnblogs.com/guizimo/p/13454977.html