[LeetCode] 1847. Closest Room

There is a hotel with n rooms. The rooms are represented by a 2D integer array rooms where rooms[i] = [roomIdi, sizei] denotes that there is a room with room number roomIdi and size equal to sizei. Each roomIdi is guaranteed to be unique.

You are also given k queries in a 2D array queries where queries[j] = [preferredj, minSizej]. The answer to the jth query is the room number id of a room such that:

• The room has a size of at least minSizej, and
• abs(id - preferredj) is minimized, where abs(x) is the absolute value of x.

If there is a tie in the absolute difference, then use the room with the smallest such id. If there is no such room, the answer is -1.

Return an array answer of length k where answer[j] contains the answer to the jth query.

Example 1:

Input: rooms = [[2,2],[1,2],[3,2]], queries = [[3,1],[3,3],[5,2]]
Output: [3,-1,3]
Explanation: The answers to the queries are as follows:
Query = [3,1]: Room number 3 is the closest as abs(3 - 3) = 0, and its size of 2 is at least 1. The answer is 3.
Query = [3,3]: There are no rooms with a size of at least 3, so the answer is -1.
Query = [5,2]: Room number 3 is the closest as abs(3 - 5) = 2, and its size of 2 is at least 2. The answer is 3.

Example 2:

Input: rooms = [[1,4],[2,3],[3,5],[4,1],[5,2]], queries = [[2,3],[2,4],[2,5]]
Output: [2,1,3]
Explanation: The answers to the queries are as follows:
Query = [2,3]: Room number 2 is the closest as abs(2 - 2) = 0, and its size of 3 is at least 3. The answer is 2.
Query = [2,4]: Room numbers 1 and 3 both have sizes of at least 4. The answer is 1 since it is smaller.
Query = [2,5]: Room number 3 is the only room with a size of at least 5. The answer is 3.

Constraints:

• n == rooms.length
• 1 <= n <= 105
• k == queries.length
• 1 <= k <= 104
• 1 <= roomIdi, preferredj <= 107
• 1 <= sizei, minSizej <= 107

最近的房间。

一个酒店里有 n 个房间，这些房间用二维整数数组 rooms 表示，其中 rooms[i] = [roomIdi, sizei] 表示有一个房间号为 roomIdi 的房间且它的面积为 sizei 。每一个房间号 roomIdi 保证是 独一无二 的。

同时给你 k 个查询，用二维数组 queries 表示，其中 queries[j] = [preferredj, minSizej] 。第 j 个查询的答案是满足如下条件的房间 id ：

房间的面积 至少 为 minSizej ，且
abs(id - preferredj) 的值 最小 ，其中 abs(x) 是 x 的绝对值。
如果差的绝对值有 相等 的，选择 最小 的 id 。如果 没有满足条件的房间 ，答案为 -1 。

请你返回长度为 k 的数组 answer ，其中 answer[j] 为第 j 个查询的结果。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/closest-room
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这道题我暂时提供一个排序 + treeset的做法。题目说的有点绕，我重新解释一下。有 N 个房间，能拿到的信息是 id 和 roomSize；对于每一个 query，也包含两个信息，[preferredId, minSize]，表示一个人 prefer 的房间号和这个人要求的最小的 roomSize。因为 queries 是一次性一起给出来的，所以我这里的思路是需要对两个 input 数组排序，尽量把 roomSize 大的房间分配给需求大的人，同时我也把roomSize大的房间尽量放在前面。

这时候我还需要一个 treeset，同时开始遍历 queries。对于每一个 query，只有当房间size满足当前这个人的 minSize，我才把 roomId 加入treeset。这样当我把满足当前这个人的 minSize 的所有房间 ID 都记录好之后，我再利用treeset的函数去找到往上和往下最接近 preferredId 的房间号。

如果对 treeset 的用法不清楚，自己查一下 floor() 和 ceiling() 这两个函数的定义。

时间O(nlogn)

空间O(n)

Java实现

class Solution {
    public int[] closestRoom(int[][] rooms, int[][] queries) {
        int[] res = new int[queries.length];
        int[][] qs = new int[queries.length][];
        for (int i = 0; i < queries.length; i++) {
            // { preferredID, minSize, i }
            qs[i] = new int[] { queries[i][0], queries[i][1], i };
        }
        // 按room size从大到小排序
        Arrays.sort(rooms, (a, b) -> b[1] - a[1]);
        // 按minSize从大到小排序
        Arrays.sort(qs, (a, b) -> b[1] - a[1]);

        int i = 0;
        TreeSet<Integer> treeset = new TreeSet<>();
        for (int[] q : qs) {
            for (; i < rooms.length && rooms[i][1] >= q[1]; i++) {
                treeset.add(rooms[i][0]);
            }
            Integer ans1 = treeset.floor(q[0]);
            Integer ans2 = treeset.ceiling(q[0]);
            if (ans1 != null && ans2 != null) {
                res[q[2]] = q[0] - ans1 <= ans2 - q[0] ? ans1 : ans2;
            } else {
                res[q[2]] = ans1 == null && ans2 == null ? -1 : ans1 == null ? ans2 : ans1;
            }
        }
        return res;
    }
}

 

LeetCode 题目总结