题目描述：23. 合并 K 个升序链表（困难）

给你一个链表数组，每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中，返回合并后的链表。

LeetCode做题链接：LeetCode-合并 K 个升序链表

示例 1：

输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[1->4->5,1->3->4,2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

示例 2：

输入：lists = []
输出：[]

示例 3：

输入：lists = [[]]
输出：[]

提示：

k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按 升序 排列
lists[i].length 的总和不超过 10^4

题目接口

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode mergeKLists(ListNode[] lists) {}
}

解题思路1

用一个变量 res 来维护以及合并的链表，第 i 次循环把第 i 个链表和 res 合并，答案保存到 res 中。

代码

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode mergeKLists(ListNode[] lists) {if (lists == null || lists.length == 0) {return null;}ListNode res = lists[0];for (int i = 1; i < lists.length; i++) {if (lists[i] == null) {continue;}res = mergeTwoLists(res, lists[i]);}return res;}private ListNode mergeTwoLists(ListNode res, ListNode list) {if (res == null) {return list;} else if (list == null) {return res;} else if (res.val < list.val) {res.next = mergeTwoLists(res.next, list);return res;} else {list.next = mergeTwoLists(res, list.next);return list;}}
}

成功！

解题思路2

可以两两有序合并的方式，然后重复两两有序合并的过程，最后得到一个有序的链表

代码

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode mergeKLists(ListNode[] lists) {return merge(lists, 0, lists.length - 1);}public ListNode merge(ListNode[] lists, int l, int r) {if (l == r) {return lists[l];}if (l > r) {return null;}int mid = (l + r) >> 1;return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));}public ListNode mergeTwoLists(ListNode a, ListNode b) {if (a == null || b == null) {return a != null ? a : b;}ListNode head = new ListNode(0);ListNode tail = head, aPtr = a, bPtr = b;while (aPtr != null && bPtr != null) {if (aPtr.val < bPtr.val) {tail.next = aPtr;aPtr = aPtr.next;} else {tail.next = bPtr;bPtr = bPtr.next;}tail = tail.next;}tail.next = (aPtr != null ? aPtr : bPtr);return head.next;}
}

成功！

PS:

感谢您的阅读！如果您觉得本篇文章对您有所帮助，请给予博主一个赞喔~