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)
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)
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)。