在Python中交换链表中的对，一个链接会消失吗

Question

我一直在学习链表，在python中实现链表比我预期的要容易。然而，当涉及到解决"在链表中交换对"的问题时，由于某种原因，我的第二个链接在交换过程中消失了。我已经盯着这个问题看了很长时间，并尝试了我在网上找到的不同解决方案。他们都得到了相同的结果，这表明我的问题在于列表本身的实现。或者我在一个我看不见的地方犯了一个愚蠢的错误！如果能有一双崭新的眼睛，我将不胜感激。我做错了什么？

class Node:
def __init__(self, val):
self.value = val
self.next = None
class LinkedList:
def __init__(self, data):
self.head = Node(data)
def printList(self, head):
while head:
print("->" , head.value)
head = head.next;
def swapPairsR(self, node): # recursive
if node is None or node.next is None:
return node
ptrOne = node
ptrTwo = node.next
nextPtrTwo = ptrTwo.next
# swap the pointers here at at the rec call
ptrTwo.next = node
newNode = ptrTwo
ptrOne.next = self.swapPairsR(nextPtrTwo)
return newNode
def swapPairsI(self, head): # iterative
prev = Node(0)
prev.next = head
temp = prev
while temp.next and temp.next.next:
ptrOne = temp.next
ptrTwo = temp.next.next
# change the pointers to the swapped pointers
temp.next = ptrTwo
ptrOne.next = ptrTwo.next
ptrTwo.next = ptrOne
temp = temp.next.next
return prev.next
thisLList = LinkedList(1)
thisLList.head.next = Node(2)
thisLList.head.next.next = Node(3)
thisLList.head.next.next.next = Node(4)
thisLList.head.next.next.next.next = Node(5)
thisLList.printList(thisLList.head)
print("--------------")
thisLList.swapPairsI(thisLList.head)
thisLList.printList(thisLList.head)

编辑：我的输出：

-> 1
-> 2
-> 3
-> 4
-> 5
--------------
-> 1
-> 4
-> 3
-> 5

Answer 1

swapPairsI函数正在返回链表的新head。你需要相应地更新它：

thisLList.head = thisLList.swapPairsI(thisLList.head)

---

或者更好的是，您应该更改swapPairsI函数，这样它就不必将节点作为参数：

def swapPairsI(self): # iterative
prev = Node(0)
prev.next = self.head
temp = prev
while temp.next and temp.next.next:
ptrOne = temp.next
ptrTwo = temp.next.next
# change the pointers to the swapped pointers
temp.next = ptrTwo
ptrOne.next = ptrTwo.next
ptrTwo.next = ptrOne
temp = temp.next.next
self.head = prev.next

在这种情况下，您可以简单地调用：

thisLList.swapPairsI()