包含每个查询的最小区间

You are given a 2D integer array intervals, where intervals[i] = [lefti, righti] describes the ith interval starting at lefti and ending at righti (inclusive). The size of an interval is defined as the number of integers it contains, or more formally righti - lefti + 1.

You are also given an integer array queries. The answer to the jth query is the size of the smallest interval i such that lefti <= queries[j] <= righti. If no such interval exists, the answer is -1.

Return an array containing the answers to the queries.

 

Example 1:

Input: intervals = [[1,4],[2,4],[3,6],[4,4]], queries = [2,3,4,5]
Output: [3,3,1,4]
Explanation: The queries are processed as follows:
- Query = 2: The interval [2,4] is the smallest interval containing 2. The answer is 4 - 2 + 1 = 3.
- Query = 3: The interval [2,4] is the smallest interval containing 3. The answer is 4 - 2 + 1 = 3.
- Query = 4: The interval [4,4] is the smallest interval containing 4. The answer is 4 - 4 + 1 = 1.
- Query = 5: The interval [3,6] is the smallest interval containing 5. The answer is 6 - 3 + 1 = 4.

Example 2:

Input: intervals = [[2,3],[2,5],[1,8],[20,25]], queries = [2,19,5,22]
Output: [2,-1,4,6]
Explanation: The queries are processed as follows:
- Query = 2: The interval [2,3] is the smallest interval containing 2. The answer is 3 - 2 + 1 = 2.
- Query = 19: None of the intervals contain 19. The answer is -1.
- Query = 5: The interval [2,5] is the smallest interval containing 5. The answer is 5 - 2 + 1 = 4.
- Query = 22: The interval [20,25] is the smallest interval containing 22. The answer is 25 - 20 + 1 = 6.

 

Constraints:

• 1 <= intervals.length <= 105
• 1 <= queries.length <= 105
• intervals[i].length == 2
• 1 <= lefti <= righti <= 107
• 1 <= queries[j] <= 107

/*
 * 思路：
 * 使用Java Stream API来实现同样的逻辑。我们可以使用Stream进行排序、过滤以及收集操作。
 * 在这里，我们可以使用Map来保存每个查询点的答案。
 */
import java.util.*;
import java.util.stream.*;

public class SolutionStream {
    public int[] minInterval(int[][] intervals, int[] queries) {
        Map<Integer, Integer> queryResults = new TreeMap<>();
        // 将queries中的查询点和原索引对应保存
        IntStream.range(0, queries.length).forEach(i -> queryResults.put(queries[i], -1));
        // 对区间按起点升序，起点相同时按终点升序排序
        Arrays.sort(intervals, Comparator.comparingInt((int[] a) -> a[0]).thenComparingInt(a -> a[1]));
        // 处理每个查询点
        PriorityQueue<int[]> minHeap = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        int index = 0;
        for (int query : queryResults.keySet()) {
            while (index < intervals.length && intervals[index][0] <= query) {
                int left = intervals[index][0];
                int right = intervals[index][1];
                minHeap.offer(new int[]{right - left + 1, right});
                index++;
            }
            while (!minHeap.isEmpty() && minHeap.peek()[1] < query) {
                minHeap.poll();
            }
            queryResults.put(query, minHeap.isEmpty() ? -1 : minHeap.peek()[0]);
        }
        // 返回结果
        return Arrays.stream(queries).map(queryResults::get).toArray();
    }
}

此题解采用排序和优先队列（堆）的方法来解决问题。首先，对查询按值进行排序，并保留原始索引以便最后能正确地放置答案。接着，对区间按起始位置进行排序。对于每个查询，使用最小堆来维护所有包含当前查询点的区间，并确保堆中区间的结束位置至少大于等于查询值。堆中存储元组，包含区间长度、起始位置和结束位置，按长度排序以快速获得最小长度的区间。对于每个查询，将所有起始位置小于等于查询值的区间加入堆中，然后移除堆中所有结束位置小于查询值的区间。如果堆非空，堆顶元素即为包含该查询值的最小区间长度。

O((n+m) log n)

O(n + m)