【问题】
Sort a linked list in O(n log n)
time using constant space complexity.
【代码】
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# @param head, a ListNode
# @return a ListNode
def sortList(self, head):
if head == None or head.next == None:
return head
fast = head
slow = head
while fast.next != None and fast.next.next != None:
fast = fast.next.next
slow = slow.next
head1 = head
head2 = slow.next
slow.next = None
head1 = self.sortList(head1)
head2 = self.sortList(head2)
head = self.merge(head1, head2)
return head
#Merge
def merge(self, head1, head2):
if head1 == None: return head2
if head2 == None: return head1
head = ListNode(0)
pre = head
while head1 != None and head2 != None:
if head1.val < head2.val:
pre.next = head1
head1 = head1.next
else:
pre.next = head2
head2 = head2.next
pre = pre.next
if head1 == None:
pre.next = head2
if head2 == None:
pre.next = head1
return head.next