23.Merge k Sorted Lists
===

###### tags: `Hard`,`Linked List`,`Heap`,`Divide and Conquer`

[23. Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)

### 題目描述

You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.

*Merge all the linked-lists into one sorted linked-list and return it.*

### 範例

**Example 1:**

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

**Example 2:**

Input: lists = []
Output: []

**Example 3:**

Input: lists = [[]]
Output: []

**Constraints**:

* `k` == `lists.length`
* 0 <= `k` <= 10^4^
* 0 <= `lists[i].length` <= 500
* -10^4^ <= `lists[i][j]` <= 10^4^
* `lists[i]` is sorted in **ascending order**.
* The sum of `lists[i].length` will not exceed 10^4^.

### 解答

#### Python

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        def merge2Lists(l1, l2):
            head = ListNode()
            cur = head
            while l1 and l2:
                if l1.val <= l2.val:
                    cur.next = l1
                    l1 = l1.next
                else:
                    cur.next = l2
                    l2 = l2.next
                cur = cur.next
            if l1: cur.next = l1
            else: cur.next = l2
            return head.next
        
        N = len(lists)
        if N <= 0: return None
        step = 1
        while step < N:
            for i in range(0, N - step, step*2):
                lists[i] = merge2Lists(lists[i], lists[i + step])
            step *= 2
        return lists[0]

> [name=Yen-Chi Chen][time=Sun, Mar 12, 2023]

### Reference

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