目錄
82. Remove Duplicates from Sorted List II
83. Remove Duplicates from Sorted List
21. Merge Two Sorted Lists
設置p1爲被插入的列表,p2是待分解的列表，則必有p1.val<=p2.val。
- p2.val<=p1.next.val則進行一次插入操作，更新p1，p2，這裏的更新操作稍微麻煩一點。
- 否則p1=p1.next
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
if l1.val>l2.val:
head1,head2=l2,l1
else:
head1,head2=l1,l2
p1,p2=head1,head2
while 1:
if p2==None:
break
elif p1 and p1.next==None:
p1.next=p2
break
if p2.val<=p1.next.val:
tmp1=p1.next
tmp2=p2.next
p1.next=p2
p2.next=tmp1
p1=p2
p2=tmp2
else:
p1=p1.next
return head1
23. Merge k Sorted Lists
可以一次merge兩個鏈表，也可以多個鏈表同時merge。
class Solution(object):
def mergeKLists(self, lists):
"""
:type lists: List[ListNode]
:rtype: ListNode
"""
def merge( l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
if l1.val>l2.val:
head1,head2=l2,l1
else:
head1,head2=l1,l2
p1,p2=head1,head2
while 1:
if p2==None:
break
elif p1 and p1.next==None:
p1.next=p2
break
if p2.val<=p1.next.val:
tmp1=p1.next
tmp2=p2.next
p1.next=p2
p2.next=tmp1
p1=p2
p2=tmp2
else:
p1=p1.next
return head1
l=len(lists)
if l==0:
return None
x=lists[0]
if l==1:
return x
else:
for i in range(1,l):
x=merge(x,lists[i])
return x
82. Remove Duplicates from Sorted List II
主要是對於head.val有重複的情況要考察一下。
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
p=head
if head==None:
return head
if head.next==None:
return head
elif head.val==head.next.val:
flag=1
else:flag=0
while p:
tmp1=p.next
if tmp1==None:
break
tmp2=tmp1.next
if tmp2==None:
break
if tmp1.val==tmp2.val:
tmp=tmp2
while tmp.val==tmp1.val:
tmp=tmp.next
if tmp==None:
break
p.next=tmp
else:
p=p.next
if flag:
if head.next==None:
return None
elif head.val==head.next.val:
return head.next.next
else:
return head.next
else:
return head
83. Remove Duplicates from Sorted List
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
p=head
while p:
tmp=p.next
if not tmp:
break
else:
if tmp.val==p.val:
if tmp.next:
p.next=tmp.next
# p=p.next
else:
p.next=None
p=None
else:
p=p.next
return head
141. Linked List Cycle
這裏採取了很笨的方法，把遍歷過的節點存起來，每次比較是否遍歷過。
也可以採取雙指針法，一個指針在前速度爲2，一個指針在後速度爲1，如果鏈表有環則兩個指針必定會相遇。
class Solution(object):
def hasCycle(self, head):
"""
:type head: ListNode
:rtype: bool
"""
p=head
visited=[]
if not head:
return False
else:
# tmp=p.next
# while tmp:
# if tmp.next==p:
# return False
# elif tmp.next==p.next:
# break
# else:
# tmp=tmp.next
# return Ture
while p:
if p in visited:
return True
else:
visited.append(p)
p=p.next
return False
328. Odd Even Linked List
要注意的是需要奇偶同步遍歷鏈表，如果分開遍歷的話，會出現不同步的問題。而同步遍歷，可能會存在最後一個odd節點沒有被遍歷到，最後判斷一下就好了。
class Solution(object):
def oddEvenList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or head.next==None or head.next.next==None:
return head
res=head
tmp=head.next
odd,even=head,head.next
# tmp=ListNode()
while(odd.next and odd.next.next and even.next and even.next.next):
odd.next=odd.next.next
even.next=even.next.next
odd=odd.next
even=even.next
# print(odd.val,even.val)
if even.next:
odd.next=even.next
odd=odd.next
print(odd.val,even.val)
even.next=None
# if odd.next!=None:
# odd=odd.next
odd.next=tmp
return res
725. Split Linked List in Parts
class Solution(object):
def splitListToParts(self, root, k):
"""
:type root: ListNode
:type k: int
:rtype: List[ListNode]
"""
p=root
l=0
while p:
l+=1
p=p.next
i=l//k
rest=l-i*k
po=[i]*k
for j in range(rest):
po[j]+=1
# print(po)
p=root
res=[]
for i in po:
if i==0:
res.append(None)
else:
res.append(p)
for j in range(i-1):
p=p.next
k=p.next
p.next=None
p=k
return res