题解 | #链表的奇偶重排#
链表的奇偶重排
http://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3
遍历两次单项链表
不使用数组作为缓存排序的方式,
class Solution:
def oddEvenList(self, head: ListNode):
h = head
res = ListNode(0)
p = res
i = 1
while head is not None:
if i % 2 != 0:
# print(head.val)
p.next = ListNode(head.val)
p = p.next
head = head.next
i += 1
head = h
while head is not None:
if i % 2 == 0:
# print(head.val)
p.next = ListNode(head.val)
p = p.next
head = head.next
i += 1
return res.next