原题
Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?
Reference Answer
思路分析
解题方法有很多种:
- 借助python list,保存每个节点,再反向遍历,重新链接即可。(时间复杂度 O(n),空间复杂度 O(n))
- 进一步优化,直接在进行遍历的时候,保存当前节点,进行反向链接即可,具体看参考代码(此时,时间复杂度O(n),空间复杂度O(1))
Code One
时间复杂度 O(n),空间复杂度 O(n)
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
temp_list = []
while head:
end = head
temp_list.append(head)
head = head.next
root = end
for index in range(len(temp_list)-2, -1, -1):
end.next = temp_list[index]
end = end.next
end.next = None
return root
Code Two
时间复杂度 O(n),空间复杂度 O(1)
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
real_end = end = head
# end.next = None
head = head.next
while head:
temp = head
head = head.next
temp.next = end
end = temp
real_end.next = None
return end
Note:
- 注意方法二中,最终要设置
real_end.next = None