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Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
AC代码：
def mergeKLists_merge_sort(self, lists): def merge(lst1, lst2): dummy = pt = ListNode(-1) while lst1 and lst2: if lst1.val < lst2.val: pt.next = lst1 lst1 = lst1.next else: pt.next = lst2 lst2 = lst2.next pt = pt.next pt.next = lst1 if not lst2 else lst2 return dummy.next if not lists: return None if len(lists) == 1: return lists[0] mid = len(lists)/2 left = self.mergeKLists_merge_sort(lists[:mid]) right = self.mergeKLists_merge_sort(lists[mid:]) return merge(left, right)
开始用的递归，发现超时了。后来借鉴了网上的代码，用的是归并排序。
此外还有优先级队列等方法，后面补充。
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原文地址：http://www.cnblogs.com/zhuifengjingling/p/5233359.html
