(LeetCode)Swap Nodes in Pairs --- 交换两个相邻的结点

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.

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解题分析：
对于此题目，分析题目，我们可以加一个索引，在第一个结点，然后可以保证第一个结点不丢失， 然后依次交换每个结点。
# -*- coding:utf-8 -*-
__author__ = 'jiuzhang'
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None

class Solution(object):
def swapPairs(self, head):
dummy = ListNode(0)
dummy.next = head
pre, cur = dummy, head
while cur and cur.next:
pre.next = cur.next
cur.next = pre.next.next
pre.next.next = cur
pre, cur = cur, cur.next
return dummy.next