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题目来源:
https://leetcode.com/problems/merge-k-sorted-lists/
题意分析：
给定k个有序的链表，将这些链表整合成一个新的有序链表。
题目思路：
前面我们已经给出了两个有序链表整合的做法。这里，我们不妨用归并排序的想法，把n个链表看成 n/2 和n - n/2的整合，直到n/2 <= 1。时间复杂度是 O(n * （2^log k）) = O(n * k).
代码（python）：
1 # Definition for singly-linked list. 2 # class ListNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution(object): 8 def mergeTwolists(self,l1,l2): 9 """ 10 :type l1: ListNode 11 :type l2: ListNode 12 :rtype: ListNode 13 """ 14 ans = ListNode(0) 15 tmp = ans 16 if l1 == None and l2 == None: 17 return None 18 while l1 !=None or l2 != None: 19 if l1 == None: 20 while l2 != None: 21 tmp.val = l2.val 22 l2 = l2.next 23 if l2 == None: 24 break 25 tmp.next = ListNode(0) 26 tmp = tmp.next 27 break 28 if l2 == None: 29 while l1 != None: 30 tmp.val = l1.val 31 l1 = l1.next 32 if l1 == None: 33 break 34 tmp.next = ListNode(0) 35 tmp = tmp.next 36 break 37 if l1.val <= l2.val: 38 tmp.val = l1.val 39 l1 = l1.next 40 else: 41 tmp.val = l2.val 42 l2 = l2.next 43 tmp.next = ListNode(0) 44 tmp = tmp.next 45 return ans 46 def mergeKLists(self, lists): 47 """ 48 :type lists: List[ListNode] 49 :rtype: ListNode 50 """ 51 size = len(lists) 52 if size == 0: 53 return None 54 if size == 1: 55 return lists[0] 56 n = size // 2 57 tmp1 = self.mergeKLists(lists[:n]) 58 tmp2 = self.mergeKLists(lists[n:]) 59 return self.mergeTwolists(tmp1,tmp2) 60
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[LeetCode]题解（python）：023-Merge k Sorted Lists
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原文地址：http://www.cnblogs.com/chruny/p/4872905.html
