链表排序-归并排序和快速排序

Posted 2020-10-15 Point

LeetCode148 SortList

题意：给定一个链表，要求用O(n log n) 的复杂度进行排序。

直观的链表归并排序：

 1 class Solution:
 2     # @param head, a ListNode
 3     # @return a ListNode
 4     def sortList(self, head):
 5         if  not head or not head.next :return head
 6         p , one,two=ListNode(0),head,head
 7         p.next=head
 8         while two and two.next:
 9             p = one
10             one , two = one.next,two.next.next
11         p.next=None  # 截断  
12         lhead=self.sortList(head)
13         rhead=self.sortList(one)
14         return self.merge(lhead,rhead)
15 
16     def merge(self,lhead,rhead):       
17         head = ListNode(-1) #表头
18         p,prep=None,head
19         while lhead and rhead:
20             if lhead.val < rhead.val:
21                 p ,lhead= lhead,lhead.next
22             else:
23                 p,rhead=rhead,rhead.next
24             prep.next=p
25             prep=prep.next
26         
27         while lhead:
28             p , lhead= lhead,lhead.next
29             prep.next=p
30             prep=prep.next
31         while rhead:
32             p,rhead=rhead,rhead.next
33             prep.next=p
34             prep=prep.next
35 
36         return head.next

 

快速排序，因为是链表不能用下标快速访问，挖坑法不适用，这里采用《算法导论》中的单向双指针法，end记录边界不采用None截断

 1 class ListNode(object):  
 2     def __init__(self, x):  
 3         self.val = x  
 4         self.next = None  
 5   
 6 class Solution(object):  
 7     def sortList(self, head):  
 8         """ 
 9         sort list using quick sort 
10         :type head: ListNode 
11         :rtype: ListNode 
12         """  
13         if head is None:  
14             return None  
15         tail = self.get_tail(head)  
16         head, tail = self.quick_sort(head, tail)  
17         tail.next = None  
18         return head  
19   
20     def quick_sort(self, head, tail):  
21         """ 
22         Sort in place 
23         :param head: 
24         :param tail: 
25         :return: 
26         """  
27         if head is not tail:  
28             head_left, tail_left, head_ref, tail_ref, head_right, tail_right = self.quicksort_partition(head, tail)  
29             if head_left is None:  # if there is no node in left part after partition  
30                 head = head_ref  
31             else:  
32                 head_left, tail_left = self.quick_sort(head_left, tail_left)  
33                 head = head_left  
34                 tail_left.next = head_ref  
35             if head_right is None:  # if there is no node in right part after partition  
36                 tail = tail_ref  
37             else:  
38                 head_right, tail_right = self.quick_sort(head_right, tail_right)  
39                 tail_ref.next = head_right  
40                 tail = tail_right  
41         return head, tail  
42   
43   
44     def quicksort_partition(self, head, tail):  
45         reference = tail  
46         head_ref, tail_ref = reference, reference  
47         head_left, tail_left, head_right, tail_right = None, None, None, None  
48   
49         sentinel = ListNode(None)  # use sentinel to simplify the code  
50         sentinel.next = head  
51         node = sentinel  
52         while node.next is not tail:  
53             node = node.next  
54             if node.val > reference.val:  # put node into right part  
55                 if head_right is not None:  
56                     tail_right.next = node  
57                     tail_right = node  
58                 else:  # right part is empty  
59                     head_right = node  
60                     tail_right = node  
61             elif node.val < reference.val:  # put node into left part  
62                 if head_left is not None:  
63                     tail_left.next = node  
64                     tail_left= node  
65                 else:  # left part is empty  
66                     head_left = node  
67                     tail_left = node  
68             else:  # put node into reference part  
69                 tail_ref.next = node  
70                 tail_ref = node  
71         return head_left, tail_left, head_ref, tail_ref, head_right, tail_right  
72   
73     def get_tail(self, node):  
74         while node.next:  
75             node = node.next  
76         return node