在 o(n log n) 時間複雜度和常數級的空間複雜度下給鍊錶排序。
給出1->3->2->null，給它排序變成1->2->3->null.
分別用歸併排序和快速排序做一遍。
實現**：
"""
definition of listnode
class listnode(object):
def __init__(self, val, next=none):
self.val = val
self.next = next
"""'''
#歸併排序
class solution:
"""@param head: the head of linked list.
@return: you should return the head of the sorted linked list, using constant space complexity.
"""def sortlist(self, head):
# write your code here
if head is none or head.next is none:
return head
#找到鍊錶的中點結點
mid = self.findmid(head)
#先執行鍊錶中點往後
right = self.sortlist(mid.next)
#將右半部分拋棄，則得到左半部分
mid.next = none
left = self.sortlist(head)
return self.mergetwolists(left, right)
def findmid(self, head):
slow, fast = head, head.next
#slow移動一位， fast移動兩位，當fast遍歷完，slow正好處於中點
while fast != none and fast.next != none:
fast = fast.next.next
slow = slow.next
return slow
def mergetwolists(self, l1, l2):
# write your code here
dummy = listnode(0)
tmp = dummy
while l1 != none and l2 != none:
if l1.val < l2.val:
tmp.next = l1
l1 = l1.next
else:
tmp.next = l2
l2 = l2.next
tmp = tmp.next
if l1 != none:
tmp.next = l1
elif l2 != none:
tmp.next = l2
return dummy.next
''''''
#快速排序
class solution:
"""@param head: the head of linked list.
@return: you should return the head of the sorted linked list, using constant space complexity.
"""def sortlist(self, head):
# write your code here
if head is none or head.next is none:
return head
mid = self.findmid(head)
leftdummy, rightdummy, middummy = listnode(0),listnode(0),listnode(0)
lefttail, righttail, midtail = leftdummy, rightdummy, middummy
while head:
#如果該結點的值小於中間結點的值，就加入左鍊錶
if head.val < mid.val:
lefttail.next = head
lefttail = lefttail.next
#如果該結點的值大於中間結點的值，就加入右鍊錶
elif head.val > mid.val:
righttail.next = head
righttail = head
#如果該結點的值等於於中間結點的值，就加入中間鍊錶
else:
midtail.next = head
midtail = head
head = head.next
#鍊錶的尾部指向none
lefttail.next, righttail.next, midtail.next = none, none, none
#對左右鍊錶開始遞迴
left = self.sortlist(leftdummy.next)
right = self.sortlist(rightdummy.next)
#合併三個鍊錶
return self.concat(left, middummy.next, right)
#找到鍊錶的中點結點
def findmid(self, head):
slow, fast = head, head.next
#slow移動一位， fast移動兩位，當fast遍歷完，slow正好處於中點
while fast != none and fast.next != none:
fast = fast.next.next
slow = slow.next
return slow
#合併三個鍊錶
def concat(self, left, mid, right):
dummy = listnode(0)
tail = dummy
tail.next = left
#找到合併後鍊錶的尾結點
tail = self.gettail(tail)
tail.next = mid
tail = self.gettail(tail)
tail.next = right
return dummy.next
#找到鍊錶的尾結點
def gettail(self, head):
if head is none:
return none
while head.next:
head = head.next
return head
'''#利用內建函式排序
class solution:
"""@param head: the head of linked list.
@return: you should return the head of the sorted linked list, using constant space complexity.
"""def sortlist(self, head):
# write your code here
from functools import cmp_to_key
nodes =
while head:
head = head.next
dummy = listnode(0)
node = dummy
for n in sorted(nodes, key=cmp_to_key(self.nodesort)):
node.next = n
node = node.next
return dummy.next
def nodesort(self, a, b):
return a.val - b.val
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