LeetCode/链表/13. 合并 K 个升序链表.md

# 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;
        
    }
}
```