Leetcode题解 23. Merge k Sorted Lists【Linked List】

题目：

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

思路：递归，判断有多少个ListNode，若只有1个则返回lists[0]，0个则返回None。其余的先sort列表的前1/2，然后sort列表的后1/2，然后排序所返回的列表。

代码：

class Solution:
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        list_num = len(lists)
        if list_num == 1:
            return lists[0]
        elif list_num == 0:
            return None
        
        list_num = list_num // 2
        list1 = self.mergeKLists(lists[:list_num])
        list2 = self.mergeKLists(lists[list_num:])
        new_list = ListNode(0)
        last = new_list
        
        while list1 != None and list2 != None:
            if list1.val < list2.val:
                last.next = list1
                list1 = list1.next
                last = last.next
            else:
                last.next = list2
                list2 = list2.next
                last = last.next
        
        if list1 == None:
            last.next = list2
        else:
            last.next = list1
        
        return new_list.next