如何实现链表的逆序

1 单向链表的反转

问题描述：

　　给定一个带头结点的单链表，请将其逆序。即如果单链表原来为head -->1 --> 2 --> 3 --> 4 --> 5，那么逆序后变为head --> 5 --> 4 --> 3 --> 2 --> 1。

解决过程：

　　给定一个单向链表1-->2-->3，通过下面的示意图，看如何一步一步的将单向列表反转。

 代码实现：

 1 class Node(object):
 2     def __init__(self, data):
 3         self.data = data
 4         self.next = None
 5 
 6 def createSingleLink():
 7     head = Node(1)
 8     cur = head
 9     for i in range(2, 10):
10         cur.next = Node(i)
11         cur = cur.next
12     return head
13 
14 def printSingleLink(head):
15     cur = head
16     while cur is not None:
17         print(cur.data, end='')
18         if cur.next is not None:
19             print("-->", end='')
20         cur = cur.next
21 
22 def Reverse(head):
23     pre = None
24     cur = head
25     while cur is not None:
26         next_ = cur.next
27         cur.next = pre
28         pre = cur
29         cur = next_
30     return pre
31 
32 if __name__ == '__main__':
33     singleHead = createSingleLink()
34     printSingleLink(singleHead)
35     reverSingleHead = Reverse(singleHead)
36     print()
37     printSingleLink(reverSingleHead)

View Code

2 双向链表反转

问题描述：

　　给定一个带头结点的双向链表，请将其逆序。即如果单链表原来为head -->1 --> 2 --> 3 --> 4 --> 5，那么逆序后变为head --> 5 --> 4 --> 3 --> 2 --> 1。

解决过程：

　　例如，给定一个带头结点的双向链表1-->2-->3，如下图看如何一步一步进行反转：

代码实现：

 1 class Node(object):
 2     def __init__(self, data):
 3         self.last = None
 4         self.data = data
 5         self.next = None
 6 
 7 def creatDoubleLink():
 8     head = Node(1)
 9     cur = head
10     for i in range(2, 10):
11         cur.next = Node(i)
12         Node(i).last = cur
13         cur = cur.next
14     return head
15 
16 def printDoubleLink(head):
17     cur = head
18     while cur:
19         print(cur.data, end='')
20         if cur.next:
21             print("-->", end='')
22         cur = cur.next
23 
24 def Reverse(head):
25     pre = None
26     cur = head
27     next_ = None
28     while cur:
29         next_ = cur.next
30         cur.next = pre
31         cur.last = next_
32         pre = cur
33         cur = next_
34     return pre
35     
36 if __name__ == '__main__':
37     doubleHead = creatDoubleLink()
38     printDoubleLink(doubleHead)
39     reveDoubleHead = Reverse(doubleHead)
40     print()
41     printDoubleLink(reveDoubleHead)

View Code

3 总结：

　　单向链表和双向链表的反转比较简单，只需做到代码一次成型，运行不出错即可。上述两种代码的实现过程，都是对原有的链表进行遍历，所以如果链表长度为N，那么它们的时间复杂度和空间复杂度为O(N)和O(1)。