题解：链表相关问题及题解（Python）
一、链表反转
1.1 头插法
class Solution:
def ReverseList(self , head: ListNode) -> ListNode:
pre = ListNode(-1)
pre.next = None
while head:
temp = head.next
head.next = pre.next
pre.next = head
head = temp
return pre.next
1.2 双指针法
class Solution:
def ReverseList(self , head: ListNode) -> ListNode:
pre = None
cur = head
while cur:
temp = cur.next
cur.next = pre
# 双指针向后移动
pre = cur
cur = temp
return pre
二、链表指定区间反转
2.1 头插法
class Solution:
def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode:
# write code here
pre = ListNode(-1)
pre.next = head
for i in range(m-1):
pre = pre.next
cur = pre.next
pre.next = None
last = cur
for i in range(m,n+1):
temp = cur.next
cur.next = pre.next
pre.next = cur
cur = temp
last.next = cur
return head
2.2 双指针法
class Solution:
def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode:
# write code here
pre = ListNode(-1)
pre.next = head
for i in range(m-1):
pre = pre.next
cur = pre.next
for i in range(m,n):
temp = cur.next
cur.next = temp.next
temp.next = pre.next
pre.next = temp
return head
三、链表每k个为一组反转
3.1 递归
class Solution:
def reverseKGroup(self , head: ListNode, k: int) -> ListNode:
# write code here
tail = head
for i in range(k):
if tail == None:
return head
else:
tail = tail.next
pre = None
cur = head
while cur != tail:
temp = cur.next
cur.next = pre
pre = cur
cur = temp
head.next = self.reverseKGroup(tail,k)
return pre
四、合并有序链表
4.1递归
class Solution:
def Merge(self , pHead1: ListNode, pHead2: ListNode) -> ListNode:
if pHead1 == None or pHead2 == None:
return pHead1 if pHead2 == None else pHead2
if pHead1.val > pHead2.val:
pHead2.next = self.Merge(pHead1, pHead2.next)
return pHead2
else:
pHead1.next = self.Merge(pHead1.next, pHead2)
return pHead1
4.2 双指针
class Solution:
def Merge(self , pHead1: ListNode, pHead2: ListNode) -> ListNode:
res = ListNode(-1)
cur = res
while pHead1 and pHead2:
if pHead1.val >= pHead2.val:
cur.next = pHead2
pHead2 = pHead2.next
else:
cur.next = pHead1
pHead1 = pHead1.next
cur = cur.next
if pHead1:
cur.next = pHead1
else:
cur.next = pHead2
return res.next
五、判断链表是否有环
5.1 追及问题（双指针）
class Solution:
def hasCycle(self , head: ListNode) -> bool:
p1 = head
p2 = head
while p2 and p2.next :
p1 = p1.next
p2 = p2.next.next
if p1 == p2:
return True
return False
5.2 哈希表
class Solution:
def hasCycle(self , head: ListNode) -> bool:
visited = set({})
p = head
while p:
if p not in visited:
visited.add(p)
p = p.next
else:
return True
return False
六、寻找入环节点
class Solution:
def hasCycle(self , head: ListNode):
p1 = head
p2 = head
while p2 and p2.next :
p1 = p1.next
p2 = p2.next.next
if p1 == p2:
return True, p1
return False, p1
def EntryNodeOfLoop(self, pHead):
# write code here
loop, p1 = self.hasCycle(pHead)
p2 = pHead
if loop ==1:
while p1 != p2:
p1 = p1.next
p2 = p2.next
return p1
else:
return None
七、链表排序
7.1 借助列表
class Solution:
def sortInList(self , head: ListNode) -> ListNode:
# write code here
p = head
list_ = []
while p:
list_.append(p.val)
p = p.next
list_ = sorted(list_,reverse=True)
res = ListNode(float('-inf'))
res.next = None
for a in list_:
node = ListNode(a)
node.next = res.next
res.next = node
return res.next
7.2 双指针+递归
class Solution:
def Merge(self, pHead1: ListNode, pHead2: ListNode):
p1 = pHead1
p2 = pHead2
res = ListNode(-1)
cur = res
while p1 and p2:
if p1.val > p2.val:
cur.next = p2
p2 = p2.next
else:
cur.next = p1
p1 = p1.next
cur = cur.next
if p1:
cur.next = p1
if p2:
cur.next = p2
return res.next
def sortInList(self , head: ListNode) -> ListNode:
# write code here
if head==None or head.next==None:
return head
fast = head.next
slow = head
while fast:
if fast.next == None:
break
slow = slow.next
fast = fast.next.next
head2 = slow.next
slow.next = None
head1 = self.sortInList(head)
head2 = self.sortInList(head2)
head = self.Merge(head1,head2)
return head
八、判断是否是回文链表
8.1 利用数组
8.2 快慢指针 + 链表逆序
class Solution:
def isPail(self , head: ListNode) -> bool:
# write code here
mid = self.Middle(head)
r_head = self.Reverse(mid)
while r_head and head:
if r_head.val != head.val:
return False
r_head = r_head.next
head = head.next
return True
def Middle(self, head):
slow = head
fast = head
while fast and fast.next and fast.next.next:
fast = fast.next.next
slow = slow.next
return slow
def Reverse(self, head):
pre = ListNode(-1)
pre.next = None
while head:
temp = head.next
head.next = pre.next
pre.next = head
head = temp
return pre.next