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‎README.adoc

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//|{doc_base_url}/0985-sum-of-even-numbers-after-queries.adoc[题解]
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//|Easy
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/interval-list-intersections/[986. Interval List Intersections^]
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//|{source_base_url}/_0986_IntervalListIntersections.java[Java]
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//|{doc_base_url}/0986-interval-list-intersections.adoc[题解]
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//|Medium
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/interval-list-intersections/[986. Interval List Intersections^]
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|{source_base_url}/_0986_IntervalListIntersections.java[Java]
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|{doc_base_url}/0986-interval-list-intersections.adoc[题解]
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|Medium
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|
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//|{counter:codes}
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//|{leetcode_base_url}/vertical-order-traversal-of-a-binary-tree/[987. Vertical Order Traversal of a Binary Tree^]
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//|{source_base_url}/_0987_VerticalOrderTraversalOfABinaryTree.java[Java]

‎docs/0986-interval-list-intersections.adoc

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[#0986-interval-list-intersections]
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= 986. Interval List Intersections
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= 986. 区间列表的交集
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{leetcode}/problems/interval-list-intersections/[LeetCode - Interval List Intersections^]
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https://leetcode.cn/problems/interval-list-intersections/[LeetCode - 986. 区间列表的交集 ^]
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Given two lists of *closed* intervals, each list of intervals is pairwise disjoint and in sorted order.
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给定两个由一些 *闭区间* 组成的列表，`firstList` 和 `secondList`，其中 `firstList[i] = [start~i~, end~i~]` 而 `secondList[j] = [start~j~, end~j~]`。每个区间列表都是成对 *不相交* 的，并且 *已经排序* 。
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Return the intersection of these two interval lists.
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返回这 *两个区间列表的交集* 。
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_(Formally, a closed interval `[a, b]` (with `a <= b`) denotes the set of real numbers `x` with `a <= x <= b`. The intersection of two closed intervals is a set of real numbers that is either empty, or can be represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].)_
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形式上，*闭区间* `[a, b]`（其中 `+a <= b+`）表示实数 `x` 的集合，而 `+a <= x <= b+`。
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两个闭区间的 *交集*是一组实数，要么为空集，要么为闭区间。例如，`[1, 3]` 和 `[2, 4]` 的交集为 `[2, 3]` 。
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*示例 1：*
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*Example 1:*
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image::images/0986-01.png[{image_attr}]
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*image::https://assets.leetcode.com/uploads/2019/01/30/interval1.png[]*
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image::images/0056-01.png[{image_attr}]
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[subs="verbatim,quotes,macros"]
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----
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*Input:* A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
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*Output:* [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
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*Reminder:* The inputs and the desired output are lists of Interval objects, and not arrays or lists.
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----
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....
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输入：firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]]
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输出：[[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
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....
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*示例 2：*
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....
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输入：firstList = [[1,3],[5,9]], secondList = []
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输出：[]
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....
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*示例 3：*
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....
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输入：firstList = [], secondList = [[4,8],[10,12]]
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输出：[]
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....
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*Note:*
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*示例 4：*
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....
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输入：firstList = [[1,7]], secondList = [[3,10]]
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输出：[[3,7]]
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....
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. `0 <= A.length < 1000`
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. `0 <= B.length < 1000`
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. `0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9`
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*提示：*
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* `+0 <= firstList.length, secondList.length <= 1000+`
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* `+firstList.length + secondList.length >= 1+`
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* `0 \<= start~i~ < end~i~ \<= 10^9^`
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* `end~i~ < start~i+1~`
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* `0 \<= start~j~ < end~j~ \<= 10^9^`
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* `end~j~ < start~j+1~`
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*NOTE:* input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
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== 思路分析
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双指针。先排除两种没有交集的情况，剩下四种有交集的情况，左边选两个里面最大的，右边选两个里面最小的。哪个小，向右移动哪个的指针。
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官方题解代码更简洁，直接左边选两个里面最大的，右边选两个里面最小的。然后再检查这两个值是否左小右大。也是“哪个小，向右移动哪个的指针。”。
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image::images/0986-10.png[{image_attr}]
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[[src-0986]]
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[tabs]
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====
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一刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0986_IntervalListIntersections.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0986_IntervalListIntersections_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/interval-list-intersections/solutions/3604/qu-jian-lie-biao-de-jiao-ji-by-leetcode/[986. 区间列表的交集 - 官方题解^]
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. https://leetcode.cn/problems/interval-list-intersections/solutions/1213336/qu-jian-lie-biao-de-jiao-ji-cyu-yan-xian-n901/[986. 区间列表的交集 - 【C语言详解】【超级详细】^]

‎docs/images/0986-01.png

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‎docs/images/0986-10.png

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‎docs/index.adoc

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// include::0985-sum-of-even-numbers-after-queries.adoc[leveloffset=+1]
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// include::0986-interval-list-intersections.adoc[leveloffset=+1]
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include::0986-interval-list-intersections.adoc[leveloffset=+1]
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// include::0987-vertical-order-traversal-of-a-binary-tree.adoc[leveloffset=+1]
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‎logbook/202503.adoc

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|{doc_base_url}/1209-remove-all-adjacent-duplicates-in-string-ii.adoc[题解]
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|✅ 暴力解法：对重复的相邻字母计数，当计数达到 k 时将其删除。重复此操作，直到没有删除的字符为止。通过 19 / 21 个测试用例。更优解：把当前字符的次数存下来，下一个字符就可以在当前字符基础上做处理，省去重复计算。官方题解提供了多种解法，非常优秀！
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|{counter:codes2503}
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|{leetcode_base_url}/interval-list-intersections/[986. 区间列表的交集^]
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|{doc_base_url}/0986-interval-list-intersections.adoc[题解]
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|✅ 双指针。先排除两种没有交集的情况，剩下四种有交集的情况，左边选两个里面最大的，右边选两个里面最小的。哪个小，向右移动哪个的指针。
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‎src/main/java/com/diguage/algo/leetcode/_0986_IntervalListIntersections.java

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package com.diguage.algo.leetcode;
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import java.util.ArrayList;
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import java.util.List;
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public class _0986_IntervalListIntersections {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-06-03 14:23:09
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*/
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public int[][] intervalIntersection(int[][] firstList, int[][] secondList) {
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if (firstList == null || secondList == null
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|| firstList.length == 0 || secondList.length == 0) {
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return new int[0][0];
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}
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int first = 0, second = 0;
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List<int[]> result = new ArrayList<>();
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while (first < firstList.length && second < secondList.length) {
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if (firstList[first][1] < secondList[second][0]) {
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first++;
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} else if (secondList[second][1] < firstList[first][0]) {
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second++;
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} else {
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int max = Math.max(firstList[first][0], secondList[second][0]);
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int min;
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if (firstList[first][1] > secondList[second][1]) {
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min = secondList[second][1];
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second++;
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} else {
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min = firstList[first][1];
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first++;
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}
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result.add(new int[]{max, min});
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}
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}
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return result.toArray(new int[][]{});
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}
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// end::answer[]
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public static void main(String[] args) {
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new _0986_IntervalListIntersections()
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.intervalIntersection(new int[][]{{0, 2}, {5, 10}, {13, 23}, {24, 25}},
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new int[][]{{1, 5}, {8, 12}, {15, 24}, {25, 26}});
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}
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}

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