leetcode 每日一题 23. 合并K个排序链表

分治法

思路：

先设计好两个链表的有序合并，然后采用分治，每次合并两个链表，知道最后得到结果

代码1：

 

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


    def merge2Lists(self, l1, l2):
        head = ListNode(-1)
        temp = head
        while l1 and l2:
            if l1.val <= l2.val:
                temp.next = l1
                l1 = l1.next
            else:
                temp.next = l2
                l2 = l2.next
            temp = temp.next
        temp.next = l1 if l1 else l2
        return head.next

 

代码2:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def mergeKLists(self, lists):
        length = len(lists)
        if length == 0:
            return None
        return self.merge(lists,0,len(lists)-1)

    def merge(self,lists, left, right):
        if left == right:
            return lists[left]
        mid = left + (right - left) // 2
        l1 = self.merge(lists, left, mid)
        l2 = self.merge(lists, mid+1, right)
        return self.merge2Lists(l1, l2)
    def merge2Lists(self, l1, l2):
        head = ListNode(-1)
        temp = head
        while l1 and l2:
            if l1.val <= l2.val:
                temp.next = l1
                l1 = l1.next
            else:
                temp.next = l2
                l2 = l2.next
            temp = temp.next
        temp.next = l1 if l1 else l2
        return head.next