iven 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.
解法：
需要建立dummy节点，（即头指针前的虚拟节点），注意在连接节点的时候，最好画个图，以免把自己搞晕了，代码如下：
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class :
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def swapPairs(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if(head == None):
return None;
dummy = ListNode(-1);
pre = dummy;
dummy.next = head;
while(pre.next and pre.next.next):
tmp = pre.next;
pre.next = tmp.next;
tmp.next = pre.next.next;
pre.next.next = tmp;
pre = tmp;
return dummy.next;
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