劍指offer刷題日記 鍊錶

鍊錶03-從尾到頭列印鍊錶（python）

#採用insert方法

class

solution
: def printlist
(self,listnode)
:if not listnode:
return
result=
while listnode:
result.
insert(0
,listnode.val)
listnode=listnode.next
return result

#採用快慢指標

class

solution
: def findkthtotail
(self,head,k)
:if not head:
return none
fast=head
slow=head
for i in
range
(k):
if fast!=none:
fast=fast.next
else
:return none
while fast:
fast=fast.next
slow=slow.next
return slow

class

solution
: def reverselist
(self,head)
:if not head:
return none
last=none
while head:
temp=head.next
head.next=last
last=head
head=temp
return last

#迭代

class

solution
: def merge
(self,phead1,phead2)
:if phead1=none:
return phead2
if phead2=none:
return phead1
pmergehead=none
if phead1.valpmergehead=phead1
pmergehead=self.
merge
(phead1.next,phead2)
else
: pmergehead=phead2
pmergehead=self.
merge
(phead1,phead2.next)
return pmergehead

#複製節點，複製random，拆分

class

solution
: def clone
(self,phead)
:if not phead:
return none
pnode=phead
while pnode:
pclone=
randomlistnode
(pnode.label)
pclone.next=pnode.next
pnode.next=pclone
pnode=pclone.next
pnode=phead
while pnode:
pclone=pnode.next
if pnode.random!=none:
pclone.random=pnode.random.next
pnode=pclone.next
pnode=phead
pclonehead=pclonenode=pnode.next
pnode.next=pclonehead.next
while pnode:
pclonenode.next=pnode.next
pclonenode=pclonenode.next
pclone.next=pclonenode.next
pnode=pnode.next
return pclonehead

class

solution
: def findfirst
(self,phead1,phead2)
:if not phead1 or not phead2:
return none
p1,p2=phead1,phead2
count_1=count_2=
0while phead1:
p1=p1.next
count_1+=
1while phead2:
p2=p2.next
count_2+=
1if count_1>count_2:
while
(count_1-count_2)
: phead1=phead1.next
count_1-=
1else
:while
(count_2-count_1)
: phead2=phead2.next
count_2-=
1while phead1 and phead2:
if phead1.val==phead2.val:
return phead1
phead1=phead1.next
phead2=phead2.next
return none

#快慢指標

class

solution
: def entrynodeofloop
(self,phead)
:if not phead or not phead.next:
return none
fast=phead.next
slow=phead
while fast!=slow and fast.next:
fast=fast.next.next
slow=slow.next
if fast=slow:
fast=phead
while fast!=slow:
fast=fast.next
slow=slow.next
return fast
return none

class

solution
: def deleteduplication
(self,phead)
: head=
listnode(0
) head.next=phead
pre=head
p=phead
while p and p.next:
if p.val==p.next.val:
while p.next and p.next.val=p.val:
p.next=p.next.next
pre.next=p.next
p=pre.next
else
: pre=p
p=p.next
return head.next

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