題目描述
合併 k 個排序連結串列，返回合併後的排序連結串列。請分析和描述演算法的複雜度。

解法1

建立一個數組，將所有元素儲存，排序之後，再連結成一個連結串列

class Solution:
   def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        
        #遍歷連結串列，將所有連結串列的所有元素放到一個數組裡面
        nodeList = []
        for
 
 i in range(len(lists)):
            currentNode = lists[i]
            #遍歷某個連結串列
            while(currentNode):
                nodeList.append(currentNode)
                currentNode = currentNode.next
                
        #根據node的val對陣列進行排序  
        nodeList = sorted(nodeList,key = lambda x:
 
x.val)
        
        #對排序好的陣列的元素，一個個地連線成新的連結串列（這裡的tempHead是為了方便設定的頭節點）
        tempHead = ListNode(0)
        currentNode = tempHead
        for i in range(len(nodeList)):
            currentNode.next = nodeList[i]
            currentNode = currentNode.next
            
        return tempHead.next
 

解法2

對lists裡面的連結串列兩兩合併，知道lists裡面只有一個元素

class Solution:
    def mergeTowLists(self, list1, list2):
        head = None
        if list1 == None and list2 == None:
            return None
        if list1 == None:
            return list2
        if list2 == None:
            return list1
        if list1.val <= list2.val:
            head = list1
            head.next = self.mergeTowLists(list1.next, list2)
        else:
            head = list2
            head.next = self.mergeTowLists(list, list2.next)
        return head
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        if lists == []:
            return None
        while len(lists) > 1:
            list1 = lists.pop()
            list2 = lists.pop()
            lists.append(self.mergeTowLists(list1, list2))
        return lists[0]

這種解法是超時的：

解法3

還是使用兩兩排序的方式，不過採用分治法的思想。
定義一個partition函式，進行二分歸併。
注意partition函式的寫法

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
class Solution:
    def mergeTowLists(self, list1, list2):
        head = None
        if list1 == None and list2 == None:
            return None
        if list1 == None:
            return list2
        if list2 == None:
            return list1
        if list1.val <= list2.val:
            head = list1
            head.next = self.mergeTowLists(list1.next, list2)
        else:
            head = list2
            head.next = self.mergeTowLists(list1, list2.next)
        return head
    def partition(self, lists, start, end):
        if start == end:
            return lists[start]
        if start < end:
            mid = (start + end) // 2
            l1 = self.partition(lists, start, mid)
            l2 = self.partition(lists, mid + 1, end)
            return self.mergeTowLists(l1, l2)
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        if lists == []:
            return None
        return self.partition(lists, 0, len(lists) - 1)

這種方法的效率要比第一種方法小點