[LeetCode]题解（python）：023-Merge k Sorted Lists

题目来源:

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

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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。

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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).

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代码（python）：

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

 class Solution(object):
     def mergeTwolists(self,l1,l2):
         """
         :type l1: ListNode
         :type l2: ListNode
         :rtype: ListNode
         """
         ans = ListNode(0)
         tmp = ans
         if l1 == None and l2 == None:
             return None
         while l1 !=None or l2 != None:
             if l1 == None:
                 while l2 != None:
                     tmp.val = l2.val
                     l2 = l2.next
                     if l2 == None:
                         break
                     tmp.next = ListNode(0)
                     tmp = tmp.next
                 break
             if l2 == None:
                 while l1 != None:
                     tmp.val = l1.val
                     l1 = l1.next
                     if l1 == None:
                         break
                     tmp.next = ListNode(0)
                     tmp = tmp.next
                 break
             if l1.val <= l2.val:
                 tmp.val = l1.val
                 l1 = l1.next
             else:
                 tmp.val = l2.val
                 l2 = l2.next
             tmp.next = ListNode(0)
             tmp = tmp.next
         return ans
     def mergeKLists(self, lists):
         """
         :type lists: List[ListNode]
         :rtype: ListNode
         """
         size = len(lists)
         if size == 0:
             return None
         if size == 1:
             return lists[0]
         n = size // 2
         tmp1 = self.mergeKLists(lists[:n])
         tmp2 = self.mergeKLists(lists[n:])
         return self.mergeTwolists(tmp1,tmp2)
         

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