合并n个已排序的链表

Merge k Sorted Lists

• 合并n个已排序的链表，新链表中的每个节点必须是来自输入的原链表的节点（即不能构造新的节点），返回新链表的头部。
• Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

example 1

input:[  3->5->8,  2->11>12,  4->8,]output:2->3->4->5->8->8->11->12

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思路

1. 参照本人之前已发表的《合并两个已排序的链表》，只需要将此算法应用n-1次即可得到新链表。

代码

# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = None    def __cmp__(self, other):        return self.val <= otherclass Solution(object):    def mergeKLists_new(self, links):        """        :type links: List[ListNode]        :rtype: ListNode        """        head = None        for i in links:            head = self.mergeTwoLists(head, i)        return head    # 为了方便阅读，给出之前的代码    # from mergeTwoLists，《合并两个已排序链表》的代码    def mergeTwoLists(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        if None in (l1, l2):            return l1 or l2        head = tail = l1 if l1.val <= l2.val else l2        a = l1 if l1.val > l2.val else l1.next        b = l2 if l1.val <= l2.val else l2.next        while a and b:            if a.val <= b.val:                tail.next = a                tail, a = tail.next, a.next            else:                tail.next = b                tail, b = tail.next, b.next        tail.next = a or b        return head

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