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From: Wonder <hezhaohui0807@163.com>
Date: Mon, 6 Oct 2025 11:03:21 +0800
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+# 合并 K 个升序链表
+
+## 题目
+
+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。
+
+
+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
+
+## 思路
+
+- 分冶
+
+## 代码
+
+```java
+/**
+ * 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) {
+        // Special
+        if (lists == null || lists.length == 0) {
+            return null;
+        }
+        // Init: To List
+        List<ListNode> nlists = new ArrayList<>(Arrays.asList(lists));
+        while (nlists.size() > 1) {
+            // Traverse: l1 + l2 = list
+            List<ListNode> tempList = new ArrayList<>();
+            for (int i = 0; i < nlists.size(); i+=2) {
+                // Get l1, l2
+                ListNode l1 = nlists.get(i);
+                ListNode l2 = null;
+                if (i + 1 < nlists.size()) {
+                    l2 = nlists.get(i + 1);
+                }
+                // Merge
+                tempList.add(mergeTwoLists(l1, l2));
+            }
+            nlists = tempList;
+        }
+        
+        return nlists.get(0);
+
+    }
+
+    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
+        // Init: DummyHead, cur
+        ListNode dummyHead = new ListNode(-1);
+        ListNode cur = dummyHead;
+        // Traverse: Compare
+        while (list1 != null && list2 != null) {
+            // val1 = list1.val;
+            // val2 = list2.val;
+            ListNode tmp = null;
+            if (list1.val > list2.val) {
+                tmp = list2;
+                list2 = list2.next;
+            } else {
+                tmp = list1;
+                list1 = list1.next;
+            }
+            cur.next = tmp;
+            cur = cur.next;
+        }
+
+        cur.next = list1 == null ? list2 : list1;
+
+        return dummyHead.next;
+        
+    }
+}
+```
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