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reorder-list.py 1.50 KB
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Allen Liu 提交于 11年前 . update
# Time: O(n)
# Space: O(1)
#
# Given a singly linked list L: L0->L1->...->Ln-1->Ln,
# reorder it to: L0->Ln->L1->Ln-1->L2->Ln-2->...
#
# You must do this in-place without altering the nodes' values.
#
# For example,
# Given {1,2,3,4}, reorder it to {1,4,2,3}.
#
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def __repr__(self):
if self:
return "{} -> {}".format(self.val, repr(self.next))
class Solution:
# @param head, a ListNode
# @return nothing
def reorderList(self, head):
if head == None or head.next == None:
return head
fast, slow, prev = head, head, None
while fast != None and fast.next != None:
fast, slow, prev = fast.next.next, slow.next, slow
current, prev.next, prev = slow, None, None
while current != None:
current.next, prev, current = prev, current, current.next
l1, l2 = head, prev
dummy = ListNode(0)
current = dummy
while l1 != None and l2 != None:
current.next, current, l1 = l1, l1, l1.next
current.next, current, l2 = l2, l2, l2.next
return dummy.next
if __name__ == "__main__":
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
head.next.next.next.next = ListNode(5)
print Solution().reorderList(head)
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