120 lines
4.0 KiB
Markdown
120 lines
4.0 KiB
Markdown
---
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id: "20220207"
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date: "2022/02/07 10:38:05"
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title: "leetcode.Q23.合并K个升序链表"
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tags: ["java", "leetcode", "链表", "多路归并"]
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index_img: https://qiniupic.fleyx.com/blog/202202071446729.png
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banner_img: https://qiniupic.fleyx.com/blog/202202071447602.jpg
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categories:
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- "算法"
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- "leetcode刷题"
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---
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## 解析思路
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leetcode 困难，题目描述[点击这里](https://leetcode-cn.com/problems/merge-k-sorted-lists/)。
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本题属于多路归并类题型。解法也是多路归并，但是怎么归呢？
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对于两个有序链表的合并可以通过直接比较链表头的元素大小，将小的拿出来放到新的链表中,但是直接比较无法用于链表数不确定的情况。对于多路归并主要有以下几种办法：
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<!-- more -->
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1. 依次归并.先让第 1 和第 2 做二路归并，将结果再和第 3 做二路归并，知道全部归并完毕
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2. 分治法.先将链表两两做二路归并，此次链表数变成 n/2(链表数为奇数时 n/2+1),然后再对结果两两做二路归并，知道最后只剩两条，继续归并即得到结果
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3. 优先队列法.类比二路归并，比较两个数中的最小值，多路归并只需要比较多个数中的最小值即可，但是如果每次都重新比较一遍会比较慢，可以通过优先队列(最小堆)这种数据结构来避免多次归并，每次从队列中取出一个节点，并将这个节点的下一个节点加入队列（如果存在）,就能避免重复比较。
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## 代码
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### 依次归并
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```java
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public ListNode so1(ListNode[] lists) {
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List<ListNode> listNodeList = Arrays.stream(lists).filter(Objects::nonNull).collect(Collectors.toList());
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if (listNodeList.isEmpty()) {
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return null;
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}
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ListNode res = listNodeList.get(0);
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for (int i = 1; i < listNodeList.size(); i++) {
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//依次和第i合并
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ListNode tempRes = new ListNode(), temp = tempRes;
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ListNode cur = listNodeList.get(i);
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while (res != null && cur != null) {
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if (res.val < cur.val) {
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temp.next = res;
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res = res.next;
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} else {
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temp.next = cur;
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cur = cur.next;
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}
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temp = temp.next;
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}
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//接上剩下的部分
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temp.next = res == null ? cur : res;
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//临时结果
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res = tempRes.next;
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}
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return res;
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}
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```
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### 分治法
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```java
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public ListNode so2(ListNode[] lists) {
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List<ListNode> listNodeList = Arrays.stream(lists).filter(Objects::nonNull).collect(Collectors.toList());
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int n = listNodeList.size();
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if (n == 0) {
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return null;
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}
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while (n > 1) {
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int count = 0;
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for (int i = 0; i + 1 < n; i += 2, count++) {
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ListNode a = listNodeList.get(i), b = listNodeList.get(i + 1), head = new ListNode(), temp = head;
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while (a != null && b != null) {
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if (a.val < b.val) {
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temp.next = a;
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a = a.next;
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} else {
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temp.next = b;
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b = b.next;
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}
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temp = temp.next;
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}
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temp.next = a == null ? b : a;
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listNodeList.set(count, head.next);
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}
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//如果为奇数，要把最后一个落单的放到count上
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if (n % 2 != 0) {
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listNodeList.set(count++, listNodeList.get(n - 1));
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}
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//新的链表数量为count
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n = count;
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}
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return listNodeList.get(0);
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}
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```
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### 优先队列
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```java
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public ListNode mergeKLists(ListNode[] lists) {
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PriorityQueue<ListNode> queue = new PriorityQueue<>(Comparator.comparingInt(a -> a.val));
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queue.addAll(Arrays.stream(lists).filter(Objects::nonNull).collect(Collectors.toList()));
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if (queue.isEmpty()) {
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return null;
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}
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ListNode res = queue.poll(), temp = res;
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while (!queue.isEmpty()) {
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if (temp.next != null) {
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queue.add(temp.next);
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}
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temp.next = queue.poll();
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temp = temp.next;
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}
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return res;
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}
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```
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