leetcode 24 Swap Nodes in Pairs

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.

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

class Solution:
# @param a ListNode
# @return a ListNode
def swapPairs(self, head):
if head==None:
return None
if head.next:
p1=head
p2=head.next
p1.next=p2.next
p2.next=head
head=p2
else:
return head
pre=p1
<span style="color:#ff0000;">if pre:</span>
p1=pre.next
<span style="color:#ff0000;">if p1:</span>
p2=p1.next
else:
p1=p2=None
while p1 and p2:
pre.next=p2
p1.next=p2.next
p2.next=p1
pre=p1
<span style="color:#ff0000;">if pre:</span>
p1=pre.next
<span style="color:#ff0000;">if p1:</span>
p2=p1.next
return head

if __name__ == '__main__':
head=ListNode(1)
head.next=ListNode(2)
head.next.next=ListNode(3)
head.next.next.next=ListNode(4)
head.next.next.next.next=ListNode(5)
p=head
while p:
print p.val,
p=p.next
print
a=Solution()
p=a.swapPairs(head)
while p:
print p.val,
p=p.next