title: LeetCode No.23
categories:
- OJ
- LeetCode
tags:
- Programing
- LeetCode
- OJ
LeetCode第二十三题
题目描述
给你一个链表数组,每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中,返回合并后的链表。
示例 1:
输入:lists = [[1,4,5],[1,3,4],[2,6]]
输出:[1,1,2,3,4,4,5,6]
解释:链表数组如下:
[
1->4->5,
1->3->4,
2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2:
输入:lists = []
输出:[]
示例 3:
输入:lists = [[]]
输出:[]
提示:
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按 升序 排列
lists[i].length 的总和不超过 10^4
代码
核心思想:采用分而治之的思想,这里合并n个链表,可以两个两个的分开合并。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
# 合并两个列表
def mergeTwoLists(self, l1, l2):
res = ListNode(None)
node = res
if l1 == None:
return l2
if l2 == None:
return l1
while l1 and l2:
if l1.val < l2.val:
node.next, l1 = l1, l1.next
else:
node.next, l2 = l2, l2.next
node = node.next
if l1:
node.next = l1
else:
node.next = l2
return res.next
# 合并多个列表
def mergeKLists(self, lists):
"""
:type lists: List[ListNode]
:rtype: ListNode
"""
if len(lists) == 0:
res = None
return res
if len(lists) == 1:
return lists[0]
init = lists[0]
for i in range(1,len(lists)):
init = self.mergeTwoLists(init,lists[i])
return init