leetcode 合并k个排序列表（三种解法）

题目描述
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。

---

解法1

建立一个数组，将所有元素保存，排序之后，再链接成一个链表

class Solution:
   def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        
        #遍历链表，将所有链表的所有元素放到一个数组里面
        nodeList = []
        for i in range(len(lists)):
            currentNode = lists[i]
            #遍历某个链表
            while(currentNode):
                nodeList.append(currentNode)
                currentNode = currentNode.next
                
        #根据node的val对数组进行排序  
        nodeList = sorted(nodeList,key = lambda x:x.val)
        
        #对排序好的数组的元素，一个个地连接成新的链表（这里的tempHead是为了方便设置的头节点）
        tempHead = ListNode(0)
        currentNode = tempHead
        for i in range(len(nodeList)):
            currentNode.next = nodeList[i]
            currentNode = currentNode.next
            
        return tempHead.next

---

解法2

对lists里面的链表两两合并，知道lists里面只有一个元素

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]

这种解法是超时的：

---

解法3

还是使用两两排序的方式，不过采用分治法的思想。
定义一个partition函数，进行二分归并。
注意partition函数的写法

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)

这种方法的效率要比第一种方法小点