【Leetcode】Merge Intervals

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题目链接：https://leetcode.com/problems/merge-intervals/

题目：

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

思路：

将区间按左端点大小进行排序
将list转化为链表，便于动态删除合并区间
遍历链表，将遍历结果返回

算法：

class LinkedNode {Interval val;LinkedNode next;public LinkedNode(Interval val, LinkedNode next) {this.val = val;this.next = next;}}public List<Interval> merge(List<Interval> intervals) {List<Interval> res = new ArrayList<Interval>();if (intervals == null || intervals.size() == 0)return res;// 按照区间的起点进行排序Collections.sort(intervals, new Comparator<Interval>() {public int compare(Interval o1, Interval o2) {if (o1.start > o2.start) {return 1;} else if (o1.start < o2.start) {return -1;} else {return 0;}}});// 改为链表结构Interval headv = new Interval(0, 0);LinkedNode head = new LinkedNode(headv, null);LinkedNode p = head;for (Interval v : intervals) {LinkedNode node = new LinkedNode(v, null);p.next = node;p = node;}// 检查、合并LinkedNode v1, v2;v1 = head.next;while (v1.next != null) {v2 = v1.next;if (v1.val.end >= v2.val.start) {// 交叉if (v1.val.end >= v2.val.end) {// v1包含v2v1.next = v2.next;} else {// 非包含的交叉v1.next = v2.next;v1.val.end = v2.val.end;}} else {v1 = v1.next;}}p = head.next;while (p != null) {res.add(p.val);p = p.next;}return res;}

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