60-合并K个排序链表-LeetCode23（python）

• 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。

• 示例

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

• 思路一

暴力方法，把所有链表的节点读入一个数组中，对数组进行排序后，转化成一个链表输出

• 代码一

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        nums = []
        #将Lists的所有节点存入nums数组中
        for ls in lists:
            h = ls 
            while h :
                nums.append(h.val)
                h =  h.next
        #对数组进行排序       
        nums.sort()
        n = len(nums)
        if n == 0:
            return
        
        #将数组重新转化为链表输出
        temp = ListNode(nums[0])
        ans = temp
        for i in range(1,n):
            temp.next = ListNode(nums[i])
            temp = temp.next
        return ans
        

• 思路二

参照官方题解，分治思想，将list两两合并，引入interval变量表示当前要合并的两个list的索引之差。每次循环之后，interval的值会变大，这是因为在合并两个list时，会把前一个list的索引下标作为合并后的list下标，而不会形成新的索引。

https://leetcode-cn.com/problems/merge-k-sorted-lists/solution/he-bing-kge-pai-xu-lian-biao-by-leetcode/

• 代码二

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        amount = len(lists)
        interval = 1 
        while interval < amount:
            for i in range(0,amount-interval,interval*2):
                lists[i] = self.merge2Lists(lists[i],lists[i+interval])
            interval *= 2
        return lists[0] if amount > 0 else None

    def merge2Lists(self,l1,l2):
        head = point = ListNode(0)
        while l1 and l2:
            if l1.val < l2.val:
                point.next = l1
                l1 = l1.next
            else:
                point.next = l2
                #l2 = l1
                #l1 = point.next.next
                l2 = l2.next
            point = point.next
        if not l1:
            point.next = l2
        else:
            point.next = l1
        return head.next