mycode

# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None class Solution(object):
def oddEvenList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
Odd = head
Even = dummyeven = head.next
while Even and Even.next:
print(Even.val,Odd.val,Odd.next.val,Odd.next.next.val)
Even.next = Even.next.next #3 6 7
print(Even.next.val,Odd.next.val,Odd.next.next.val)
Even = Even.next
Odd.next = Odd.next.next # 5 4 Null
print(Odd.next.val,Odd.next.val,Odd.next.next.val)
Odd = Odd.next
print(Even.val,Odd.val)
Odd.next = dummyeven #因为无论even=None还是Even.next=None,其实在没有Even = Even.next之前,Even.next=head链中它的下一个,所以不用分类讨论加不加None return head

错误原因:Even节点变化之后,Odd取next和.next.next时,链表都已经变化了

参考:保证链表取值过程中相对顺序不乱!

# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None class Solution(object):
def oddEvenList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head:
return head
Odd = head
Even = dummyeven = head.next
while Even and Even.next:
Odd.next = Even.next
Odd = Odd.next
Even.next = Odd.next #3 6 7
Even = Even.next
Odd.next = dummyeven #因为无论even=None还是Even.next=None,其实在没有Even = Even.next之前,Even.next=head链中它的下一个,所以不用分类讨论加不加None
return head

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