A concert hall has n rows numbered from 0 to n - 1, each with m seats, numbered from 0 to m - 1. You need to design a ticketing system that can allocate seats in the following cases:
k spectators can sit together in a row.k spectators can get a seat. They may or may not sit together.Note that the spectators are very picky. Hence:
\n\nmaxRow. maxRow can vary from group to group.Implement the BookMyShow class:
BookMyShow(int n, int m) Initializes the object with n as number of rows and m as number of seats per row.int[] gather(int k, int maxRow) Returns an array of length 2 denoting the row and seat number (respectively) of the first seat being allocated to the k members of the group, who must sit together. In other words, it returns the smallest possible r and c such that all [c, c + k - 1] seats are valid and empty in row r, and r <= maxRow. Returns [] in case it is not possible to allocate seats to the group.boolean scatter(int k, int maxRow) Returns true if all k members of the group can be allocated seats in rows 0 to maxRow, who may or may not sit together. If the seats can be allocated, it allocates k seats to the group with the smallest row numbers, and the smallest possible seat numbers in each row. Otherwise, returns false.\n
Example 1:
\n\n\nInput\n["BookMyShow", "gather", "gather", "scatter", "scatter"]\n[[2, 5], [4, 0], [2, 0], [5, 1], [5, 1]]\nOutput\n[null, [0, 0], [], true, false]\n\nExplanation\nBookMyShow bms = new BookMyShow(2, 5); // There are 2 rows with 5 seats each \nbms.gather(4, 0); // return [0, 0]\n // The group books seats [0, 3] of row 0. \nbms.gather(2, 0); // return []\n // There is only 1 seat left in row 0,\n // so it is not possible to book 2 consecutive seats. \nbms.scatter(5, 1); // return True\n // The group books seat 4 of row 0 and seats [0, 3] of row 1. \nbms.scatter(5, 1); // return False\n // There is only one seat left in the hall.\n\n\n
\n
Constraints:
\n\n1 <= n <= 5 * 1041 <= m, k <= 1090 <= maxRow <= n - 15 * 104 calls in total will be made to gather and scatter.一个音乐会总共有 n 排座位，编号从 0 到 n - 1 ，每一排有 m 个座椅，编号为 0 到 m - 1 。你需要设计一个买票系统，针对以下情况进行座位安排：
k 位观众坐在 同一排座位，且座位连续 。k 位观众中 每一位 都有座位坐，但他们 不一定 坐在一起。由于观众非常挑剔，所以：
\n\nmaxRow ，这个组才能订座位。每一组的 maxRow 可能 不同 。请你实现 BookMyShow 类：
BookMyShow(int n, int m) ，初始化对象，n 是排数，m 是每一排的座位数。int[] gather(int k, int maxRow) 返回长度为 2 的数组，表示 k 个成员中 第一个座位 的排数和座位编号，这 k 位成员必须坐在 同一排座位，且座位连续 。换言之，返回最小可能的 r 和 c 满足第 r 排中 [c, c + k - 1] 的座位都是空的，且 r <= maxRow 。如果 无法 安排座位，返回 [] 。boolean scatter(int k, int maxRow) 如果组里所有 k 个成员 不一定 要坐在一起的前提下，都能在第 0 排到第 maxRow 排之间找到座位，那么请返回 true 。这种情况下，每个成员都优先找排数 最小 ，然后是座位编号最小的座位。如果不能安排所有 k 个成员的座位，请返回 false 。\n\n
示例 1：
\n\n\n输入：\n[\"BookMyShow\", \"gather\", \"gather\", \"scatter\", \"scatter\"]\n[[2, 5], [4, 0], [2, 0], [5, 1], [5, 1]]\n输出：\n[null, [0, 0], [], true, false]\n\n解释：\nBookMyShow bms = new BookMyShow(2, 5); // 总共有 2 排，每排 5 个座位。\nbms.gather(4, 0); // 返回 [0, 0]\n // 这一组安排第 0 排 [0, 3] 的座位。\nbms.gather(2, 0); // 返回 []\n // 第 0 排只剩下 1 个座位。\n // 所以无法安排 2 个连续座位。\nbms.scatter(5, 1); // 返回 True\n // 这一组安排第 0 排第 4 个座位和第 1 排 [0, 3] 的座位。\nbms.scatter(5, 1); // 返回 False\n // 总共只剩下 1 个座位。\n\n\n
\n\n
提示：
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"book": null,
"isSubscribed": false,
"isDailyQuestion": false,
"dailyRecordStatus": null,
"editorType": "CKEDITOR",
"ugcQuestionId": null,
"style": "LEETCODE",
"exampleTestcases": "[\"BookMyShow\",\"gather\",\"gather\",\"scatter\",\"scatter\"]\n[[2,5],[4,0],[2,0],[5,1],[5,1]]",
"__typename": "QuestionNode"
}
}
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