在Python的链表中交换对，一个链接消失了吗? [英] Swapping pairs in a linked list in Python, one link disappears?

问题描述

我一直在学习链表，并且在python中实现一个比我预期的要容易.但是，当解决交换链表中的对"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.多年来，我一直在盯着这个，并尝试在网上找到的其他解决方案.它们都得到相同的结果，这表明我的问题在于列表本身的实现.或者我在一个看不见的地方犯了一个愚蠢的错误！我将感谢您的新一双眼睛.我做错了什么?

I have been learning about linked lists, and implementing one in python has been easier than I expected. However, when it has come to solving the problem of "swapping pairs in a linked list", for some reason my second link disappears during the swap. I've been staring at this for ages and trying different solutions I found online. They all get the same result which suggests my problem is with the implementation of the list itself. Or I've made a silly error somewhere that I can't see! I would be grateful for a fresh pair of eyes. What have I done wrong?

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)

我的输出:

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

推荐答案

您的 swapPairsI 函数返回链表的新 head .您需要相应地更新它:

Your swapPairsI function is returning the new head of your linked list. You need to update it accordingly:

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

---

或者更好的是，您应该更改您的 swapPairsI 函数，以使其不必将节点作为参数:

---

Or better yet, you should change your swapPairsI function so that it doesn't have to take a node as parameter:

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，您可以简单地致电:

In which case, you can simply call:

thisLList.swapPairsI()

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