Leetcode题库——23.合并k个排序链表

---

@author: ZZQ
@software: PyCharm
@file: mergeKLists.py
@time: 2018/10/12 19:55
说明：合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。
示例 ：
输入:
[
1->4->5,
1->3->4,
2->6
]
输出: 1->1->2->3->4->4->5->6
思路：两两合并再合并，判断奇偶，每次用一个新的数组来存放当前经过合并后的新的链表首节点，时间复杂度：O(nlogn)

# 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
        """
        if l1 is None and l2 is None:
            return None

        if l1 is None:
            return l2

        if l2 is None:
            return l1

        l3 = ListNode(0)
        head = l3
        while l1 is not None and l2 is not None:
            if l1.val > l2.val:
                l3.next = l2
                l2 = l2.next
            else:
                l3.next = l1
                l1 = l1.next
            l3 = l3.next
        if l1 is not None and l2 is None:
            l3.next = l1
        if l1 is None and l2 is not None:
            l3.next = l2

        return head.next

    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        if len(lists) == 0:
            return None
        cur_list = lists
        while len(cur_list) > 1:
            cur_temp_list = []
            if len(cur_list) % 2:
                for i in range((len(cur_list) - 1) / 2):
                    cur_temp_list.append(self.mergeTwoLists(cur_list[i * 2], cur_list[i * 2 + 1]))
                cur_temp_list.append(cur_list[len(cur_list)-1])
            else:
                for i in range(len(cur_list) / 2):
                    cur_temp_list.append(self.mergeTwoLists(cur_list[i * 2], cur_list[i * 2 + 1]))
            cur_list = cur_temp_list
        return cur_list[0]