mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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183 lines
24 KiB
JSON
183 lines
24 KiB
JSON
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"title": "Find Winner on a Tic Tac Toe Game",
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"content": "<p><strong>Tic-tac-toe</strong> is played by two players <code>A</code> and <code>B</code> on a <code>3 x 3</code> grid. The rules of Tic-Tac-Toe are:</p>\n\n<ul>\n\t<li>Players take turns placing characters into empty squares <code>' '</code>.</li>\n\t<li>The first player <code>A</code> always places <code>'X'</code> characters, while the second player <code>B</code> always places <code>'O'</code> characters.</li>\n\t<li><code>'X'</code> and <code>'O'</code> characters are always placed into empty squares, never on filled ones.</li>\n\t<li>The game ends when there are <strong>three</strong> of the same (non-empty) character filling any row, column, or diagonal.</li>\n\t<li>The game also ends if all squares are non-empty.</li>\n\t<li>No more moves can be played if the game is over.</li>\n</ul>\n\n<p>Given a 2D integer array <code>moves</code> where <code>moves[i] = [row<sub>i</sub>, col<sub>i</sub>]</code> indicates that the <code>i<sup>th</sup></code> move will be played on <code>grid[row<sub>i</sub>][col<sub>i</sub>]</code>. return <em>the winner of the game if it exists</em> (<code>A</code> or <code>B</code>). In case the game ends in a draw return <code>"Draw"</code>. If there are still movements to play return <code>"Pending"</code>.</p>\n\n<p>You can assume that <code>moves</code> is valid (i.e., it follows the rules of <strong>Tic-Tac-Toe</strong>), the grid is initially empty, and <code>A</code> will play first.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/09/22/xo1-grid.jpg\" style=\"width: 244px; height: 245px;\" />\n<pre>\n<strong>Input:</strong> moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]\n<strong>Output:</strong> "A"\n<strong>Explanation:</strong> A wins, they always play first.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/09/22/xo2-grid.jpg\" style=\"width: 244px; height: 245px;\" />\n<pre>\n<strong>Input:</strong> moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]\n<strong>Output:</strong> "B"\n<strong>Explanation:</strong> B wins.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/09/22/xo3-grid.jpg\" style=\"width: 244px; height: 245px;\" />\n<pre>\n<strong>Input:</strong> moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]\n<strong>Output:</strong> "Draw"\n<strong>Explanation:</strong> The game ends in a draw since there are no moves to make.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= moves.length <= 9</code></li>\n\t<li><code>moves[i].length == 2</code></li>\n\t<li><code>0 <= row<sub>i</sub>, col<sub>i</sub> <= 2</code></li>\n\t<li>There are no repeated elements on <code>moves</code>.</li>\n\t<li><code>moves</code> follow the rules of tic tac toe.</li>\n</ul>\n",
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"translatedTitle": "找出井字棋的获胜者",
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"translatedContent": "<p><em>A</em> 和 <em>B</em> 在一个 <em>3</em> x <em>3</em> 的网格上玩井字棋。</p>\n\n<p>井字棋游戏的规则如下：</p>\n\n<ul>\n\t<li>玩家轮流将棋子放在空方格 (" ") 上。</li>\n\t<li>第一个玩家 A 总是用 "X" 作为棋子，而第二个玩家 B 总是用 "O" 作为棋子。</li>\n\t<li>"X" 和 "O" 只能放在空方格中，而不能放在已经被占用的方格上。</li>\n\t<li>只要有 3 个相同的（非空）棋子排成一条直线（行、列、对角线）时，游戏结束。</li>\n\t<li>如果所有方块都放满棋子（不为空），游戏也会结束。</li>\n\t<li>游戏结束后，棋子无法再进行任何移动。</li>\n</ul>\n\n<p>给你一个数组 <code>moves</code>，其中每个元素是大小为 <code>2</code> 的另一个数组（元素分别对应网格的行和列），它按照 <em>A</em> 和 <em>B</em> 的行动顺序（先 <em>A</em> 后 <em>B</em>）记录了两人各自的棋子位置。</p>\n\n<p>如果游戏存在获胜者（<em>A</em> 或 <em>B</em>），就返回该游戏的获胜者；如果游戏以平局结束，则返回 "Draw"；如果仍会有行动（游戏未结束），则返回 "Pending"。</p>\n\n<p>你可以假设 <code>moves</code> 都 <strong>有效</strong>（遵循井字棋规则），网格最初是空的，<em>A</em> 将先行动。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]\n<strong>输出：</strong>"A"\n<strong>解释：</strong>"A" 获胜，他总是先走。\n"X " "X " "X " "X " "<strong>X</strong> "\n" " -> " " -> " X " -> " X " -> " <strong>X</strong> "\n" " "O " "O " "OO " "OO<strong>X</strong>"\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]\n<strong>输出：</strong>"B"\n<strong>解释：</strong>"B" 获胜。\n"X " "X " "XX " "XXO" "XXO" "XX<strong>O</strong>"\n" " -> " O " -> " O " -> " O " -> "XO " -> "X<strong>O</strong> " \n" " " " " " " " " " "<strong>O</strong> "\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]\n<strong>输出：</strong>"Draw"\n<strong>输出：</strong>由于没有办法再行动，游戏以平局结束。\n"XXO"\n"OOX"\n"XOX"\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre><strong>输入：</strong>moves = [[0,0],[1,1]]\n<strong>输出：</strong>"Pending"\n<strong>解释：</strong>游戏还没有结束。\n"X "\n" O "\n" "\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= moves.length <= 9</code></li>\n\t<li><code>moves[i].length == 2</code></li>\n\t<li><code>0 <= moves[i][j] <= 2</code></li>\n\t<li><code>moves</code> 里没有重复的元素。</li>\n\t<li><code>moves</code> 遵循井字棋的规则。</li>\n</ul>\n",
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