24. Swap Nodes in Pairs

teminusign

于 2016-07-26 19:35:57 发布

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本文介绍了一种在常数空间复杂度下交换链表中每两个相邻节点的算法，并提供了完整的Python实现代码。该算法不修改节点值，仅改变节点之间的连接关系。

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24. Swap Nodes in Pairs

Leetcode link for this question

Discription:

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

Analyze:

Code 1 :

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """

        link=ListNode(0)
        link.next=head
        tail=link
        while tail.next and tail.next.next:
            a=tail.next
            b=a.next
            tail.next,a.next,b.next=b,b.next,a
            tail=a

        return link.next