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[leetcode]Swap Nodes in Pairs @ Python
原题地址：http://oj.leetcode.com/problems/swap-nodes-in-pairs/
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.
题意：将链表中的节点两两交换。Given 1->2->3->4, you should return the list as 2->1->4->3.
解题思路：这题主要涉及到链表的节本操作。加一个头结点，操作起来会很方便。另外配了一个示意图 ［本图是我asrman原创］
代码：
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: # @param a ListNode # @return a ListNode def swapPairs(self, head): pre = ListNode(0) pre.next = head curr = head head = pre while curr and curr.next: # curr =1, curr.next =2 pre.next = curr.next # 0 --> 2 curr.next = pre.next.next # 1 --> 3 # curr.next.next pre.next.next = curr # 3 --> 1 pre = curr # pre = 1 curr = curr.next # curr= 3 return head.next
参考：
https://oj.leetcode.com/discuss/3608/seeking-for-a-better-solution
[leetcode]Swap Nodes in Pairs @ Python
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