链表问题（3）-----反转

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)

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二、题目：反转部分单链表

 思路：

简单来说，找到要反转的部分反转，然后将前面部分和后面部分连接起来即可。

代码：

class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
def reverseList(head,a,b):
    if not head:
        return head
    n = 0
    pre = Node(None)
    while head:
        n += 1
        pre = head if n == a-1  else pre
        head = head.next
    if a > b or b > n or a < 1: 
        return head
    head ,first = pre.next , pre.next
    while a < b:
        tail = head.next.next
        temp = head.next
        head.next = tail
        temp.next = first
        first = temp
        a += 1
    if pre:
        pre.next = first
    return pre

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)
a = 2
b = 4
reverseList(head,a,b)

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三、将单链表的每k个节点之间逆序

法一：时间O（N），空间O（K）。利用栈结构存k个数

法二：时间O（N），空间O（1）