链表问题-----反转

Posted 2021-01-09 lee-yl

1、题目：反转单链表或双链表

要求：如果链表长度为N，时间复杂度为O（N），额外的空间复杂度为O（1）

反转单链表的思路：

1 → 2 → 3 → 4 → 5

（1）first = head = 1  

循环：

　　 temp = head.next   2

        head.next = temp.next   1 → 3

        temp.next = first    2 →1

        first = temp            2    【此时的头结点是2】

代码：

class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
def reverseList(head):
    if not head:
        return head
    first = head
    while head.next:
        temp = head.next
        head.next = head.next.next
        temp.next = first
        first = temp
    return first
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
reverseList(head)

反转双链表的思路：

只要将每个节点的 next 和 last 互换就可以了。

代码：

class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
def reverseList(head):
    if not head:
        return head

#重点
    while head:
        head.next , head.last = head.last,head.next
        head = head.last
    return head


head = Node(1)
head.last = None
head.next = Node(2)
head.next.last = head
head.next.next = Node(3)
head.next.next.last = head.next
head.next.next.next = Node(4)
head.next.next.next.last = head.next.next
head.next.next.next.next = Node(5)
head.next.next.next.next.last =  head.next.next.next
reverseList(head)