目录
206反转链表【链表结构基础】 21合并两个有序链表【递归】 我的答案【错误】 自己修改【超出时间限制】 在官方那里学到的【然后自己复写,错误】 对照官方【自己修改】 160相交链表【未理解题目目的】 在b站up那里学到的【然后自己复写,错误】 【超出时间限制】 对照官方【自己修改】 19删除链表的倒数第 N 个结点 看完官方解题思路【自己复写，错误】 报错信息 第一遍修改 第二遍修改 148排序链表 我的答案【冒泡排序：最小的负数输出时变为0】
class Solution ( object ) :
def reverseList ( self, head) :
"""
:type head: ListNode
:rtype: ListNode
"""
pre= None
cur= head
while cur!= None :
temp= cur. next
cur. next = pre
pre= cur
cur= temp
return pre
class Solution ( object ) :
def mergeTwoLists ( self, list1, list2) :
"""
:type list1: Optional[ListNode]
:type list2: Optional[ListNode]
:rtype: Optional[ListNode]
"""
resol= ListNode( )
while list1!= None and list2!= None :
if list1. val<= list2. val:
resol. next = ListNode( list1. val)
else :
resol. next = ListNode( list2. val)
list1= list1. next
list2= list2. next
while list1!= None :
resol. next = ListNode( list1. val)
while list2!= None :
resol. next = ListNode( list2. val)
return resol. next
class Solution ( object ) :
def mergeTwoLists ( self, list1, list2) :
"""
:type list1: Optional[ListNode]
:type list2: Optional[ListNode]
:rtype: Optional[ListNode]
"""
resol= ListNode( 0 )
temp= resol
while list1 and list2:
if list1. val<= list2. val:
resol. next = list1
list1= list1. next
else :
resol. next = list2
list2= list2. next
resol= resol. next
while list1:
resol. next = list1
while list2:
resol. next = list2
return temp. next
class Solution ( object ) :
def mergeTwoLists ( self, list1, list2) :
"""
:type list1: Optional[ListNode]
:type list2: Optional[ListNode]
:rtype: Optional[ListNode]
"""
if list1 is None :
return list2
elif list2 is None :
return list1
else :
if list1. val< list2. val:
list1. next = mergeTwoLists( self, list1. next , list2)
return list1
else :
list2. next = mergeTwoLists( self, list2. next , list1)
return list2
class Solution ( object ) :
def mergeTwoLists ( self, list1, list2) :
"""
:type list1: Optional[ListNode]
:type list2: Optional[ListNode]
:rtype: Optional[ListNode]
"""
if list1 is None :
return list2
elif list2 is None :
return list1
elif list1. val< list2. val:
list1. next = self. mergeTwoLists( list1. next , list2)
return list1
else :
list2. next = self. mergeTwoLists( list2. next , list1)
return list2
我看输入部分都已经给了要输出的结果了，就不明白这题目的是要做什么。 class Solution ( object ) :
def getIntersectionNode ( self, headA, headB) :
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
p= headA, q= headB
while ( p!= q) :
p ? p. next : headB
q ? q. next : headA
print ( "p.val" )
class Solution ( object ) :
def getIntersectionNode ( self, headA, headB) :
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
p= headA
q= headB
while ( p!= q) :
if p. next != None :
p= p. next
else :
p= headB
if q. next != None :
q= q. next
else :
q= headA
return p
class Solution ( object ) :
def getIntersectionNode ( self, headA, headB) :
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
p= headA
q= headB
while ( p!= q) :
p = p. next if p else headB
q = q. next if q else headA
return p
class Solution ( object ) :
def removeNthFromEnd ( self, head, n) :
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
p= head
for i in range ( n+ 1 ) :
p= p. next
q= head
while p!= None :
q= q. next
p= p. next
q. next = q. next . next
return head
查阅资料后发现：【self.head是指向第一个元素的，此时为None,所以没有next属性，自然就会报错。因此第一个结点需要特殊处理，我们一般通过增加头结点的方式来避免这种特殊处理。 】 发现当链表有n个元素时，无法删除倒数第n个元素。 【没有考虑到第一个结点的特殊性。】 class Solution ( object ) :
def removeNthFromEnd ( self, head, n) :
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
p= ListNode( 0 , head)
for i in range ( n+ 1 ) :
p= p. next
q= ListNode( 0 , head)
while p!= None :
q= q. next
p= p. next
q. next = q. next . next
return head
class Solution ( object ) :
def removeNthFromEnd ( self, head, n) :
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
dummy= ListNode( 0 , head)
p= dummy
for i in range ( n+ 1 ) :
p= p. next
q= dummy
while p!= None :
q= q. next
p= p. next
q. next = q. next . next
return dummy. next
class Solution ( object ) :
def sortList ( self, head) :
"""
:type head: ListNode
:rtype: ListNode
"""
dummy= ListNode( 0 , head)
q= dummy
while q. next != None :
p= dummy
while p. next != None :
if p. val> p. next . val:
temp= p. val
p. val= p. next . val
p. next . val= temp
p= p. next
q= q. next
return dummy. next
class Solution ( object ) :
def sortList ( self, head) :
def sortFunc ( head, tail) :
if not head:
return head
if head. next == tail:
head. next = None
return head
slow = fast = head
while fast != tail:
slow = slow. next
fast = fast. next
if fast != tail:
fast = fast. next
mid = slow
return merge( sortFunc( head, mid) , sortFunc( mid, tail) )
def merge ( head1, head2) :
dummyHead = ListNode( 0 )
temp, temp1, temp2 = dummyHead, head1, head2
while temp1 and temp2:
if temp1. val <= temp2. val:
temp. next = temp1
temp1 = temp1. next
else :
temp. next = temp2
temp2 = temp2. next
temp = temp. next
if temp1:
temp. next = temp1
elif temp2:
temp. next = temp2
return dummyHead. next
return sortFunc( head, None )
