***148. Sort List 排序链表

1. 原始题目

在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序。

示例 1:

输入: 4->2->1->3
输出: 1->2->3->4

示例 2:

输入: -1->5->3->4->0
输出: -1->0->3->4->5

2. 思路

归并排序。这道题两个递归解决。非常简洁。快慢指针+归并排序。

3. 解题

 1 class Solution:
 2     def sortList(self, head: ListNode) -> ListNode:
 3         if not head or not head.next:return head
 4         p,q = head, head.next
 5         while(q and q.next):          # 快慢指针，p最终指向中点
 6             p = p.next
 7             q = q.next.next
 8         head2 = p.next
 9         p.next = None
10         head = self.sortList(head)    # 排好序的左链表
11         head2 = self.sortList(head2)   # 排好序的右链表
12         head = self.merge(head,head2)   # 合并两个排好序的链表
13         return head
14         
15     def merge(self,head1,head2):   # 递归merge两个排好序的链表
16         if not head1:return head2
17         if not head2:return head1
18         if head1.val<head2.val:
19             next = head1.next
20             head1.next = self.merge(next,head2)
21             return head1
22         else:
23             next = head2.next
24             head2.next = self.merge(next,head1)
25             return head2

 

leetcode上有说可以将结点的所有数都保存下来再排序。排好序后再放到链表节点中。这种做法快，但并未实质性地交换结点。也可作为一种备选。

 1 # Definition for singly-linked list.
 2 # class ListNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution:
 8     def sortList(self, head: 'ListNode') -> 'ListNode':
 9         x=[];
10         temp=head;
11         while(head != None):
12             x.append(head.val);
13             head=head.next;
14         x.sort();
15         head =temp;
16         for i in x:
17             temp.val=i;
18             temp=temp.next;
19         return head;

 

来源：https://www.cnblogs.com/king-lps/p/10848182.html