Leetcode題庫——23.合并k個排序連結清單

@author: ZZQ

@software: PyCharm

@file: mergeKLists.py

@time: 2018/10/12 19:55

說明：合并 k 個排序連結清單，傳回合并後的排序連結清單。請分析和描述算法的複雜度。

示例 ：

輸入:

[

1->4->5,

1->3->4,

2->6

]

輸出: 1->1->2->3->4->4->5->6

思路：兩兩合并再合并，判斷奇偶，每次用一個新的數組來存放目前經過合并後的新的連結清單首節點，時間複雜度：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]
        

           

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