python - 在 Python 的链表中交换对，一个链接消失了？

Question

我一直在学习链表，用 Python 实现链表比我预期的要容易。但是，当解决“在链表中交换对”的问题时，出于某种原因，我的第二个链接在交换过程中消失了。我一直盯着这个看很久，并尝试了我在网上找到的不同解决方案。他们都得到相同的结果，这表明我的问题出在列表本身的实现上。或者我在某个地方犯了一个我看不到的愚蠢错误!如果有一双新鲜的眼睛，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

您的 swapPairsI 函数正在返回链表的新 head。您需要相应地更新它:

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()