23. Merge k Sorted Lists-优快云博客

本文链接：https://blog.youkuaiyun.com/scut_salmon/article/details/90020479

本文介绍了一种高效解决LeetCode上“合并K个有序链表”问题的方法，通过使用分治算法，将复杂度优化至O(nlogk)，避免了两两合并时可能出现的超时情况。

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https://leetcode.com/problems/merge-k-sorted-lists/

思路

一个简单的思路是把k个数组两两相加，但这样会超时，所以采用分治算法。合并两个数组的复杂度是O(n)的话，总的复杂度应该是O(nlog k)

AC代码

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        return self.merge2Lists(lists)
    def merge2Lists(self, lists):
        if len(lists) == 0:
            return None
        if len(lists) == 1:
            return lists[0]
        if len(lists) > 2:
            L = len(lists)
            return self.merge2Lists([self.merge2Lists(lists[:L/2]), self.merge2Lists(lists[L/2:])])
        l1, l2= lists
        if l1 == None:
            return l2
        if l2 == None:
            return l1
        res = ListNode(0)
        head = res
        while(l1 and l2):
            if l1.val < l2.val:
                res.val = l1.val
                res.next = ListNode(0)
                res = res.next
                l1 = l1.next
            else:
                res.val = l2.val
                res.next = ListNode(0)
                res = res.next
                l2 = l2.next
        if l1:
            res.val = l1.val
            res.next = l1.next
        else:
            res.val = l2.val
            res.next = l2.next
        return head