mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-25 17:50:26 +08:00
193 lines
20 KiB
JSON
193 lines
20 KiB
JSON
{
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"questionId": "1000374",
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"questionFrontendId": "LCP 48",
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"title": "无限棋局",
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"titleSlug": "fsa7oZ",
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"content": "English description is not available for the problem. Please switch to Chinese.",
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"translatedTitle": "无限棋局",
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"translatedContent": "小力正在通过残局练习来备战「力扣挑战赛」中的「五子棋」项目，他想请你能帮他预测当前残局的输赢情况。棋盘中的棋子分布信息记录于二维数组 `pieces` 中，其中 `pieces[i] = [x,y,color]` 表示第 `i` 枚棋子的横坐标为 `x`，纵坐标为 `y`，棋子颜色为 `color`(`0` 表示黑棋，`1` 表示白棋)。假如黑棋先行，并且黑棋和白棋都按最优策略落子，请你求出当前棋局在三步（按 **黑、白、黑** 的落子顺序）之内的输赢情况（三步之内先构成同行、列或对角线连续同颜色的至少 5 颗即为获胜）：\n- 黑棋胜, 请返回 `\"Black\"`\n- 白棋胜, 请返回 `\"White\"`\n- 仍无胜者, 请返回 `\"None\"`\n\n**注意：** \n- 和传统的五子棋项目不同，「力扣挑战赛」中的「五子棋」项目 **不存在边界限制**，即可在 **任意位置** 落子；\n- 黑棋和白棋均按 3 步内的输赢情况进行最优策略的选择\n- 测试数据保证所给棋局目前无胜者；\n- 测试数据保证不会存在坐标一样的棋子。\n\n**示例 1：**\n> 输入：\n> `pieces = [[0,0,1],[1,1,1],[2,2,0]]`\n>\n> 输出：`\"None\"`\n>\n> 解释：无论黑、白棋以何种方式落子，三步以内都不会产生胜者。\n\n**示例 2：**\n> 输入：\n> `pieces = [[1,2,1],[1,4,1],[1,5,1],[2,1,0],[2,3,0],[2,4,0],[3,2,1],[3,4,0],[4,2,1],[5,2,1]]`\n>\n> 输出：`\"Black\"`\n>\n> 解释：三步之内黑棋必胜，以下是一种可能的落子情况：\n>{:width=\"300px\"}\n\n\n\n**提示：**\n- `0 <= pieces.length <= 1000`\n- `pieces[i].length = 3`\n- `-10^9 <= pieces[i][0], pieces[i][1] <=10^9` \n- `0 <= pieces[i][2] <=1`\n",
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