给你一个链表,两两交换其中相邻的节点,并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题(即,只能进行节点交换)。
示例 1:
输入:head = [1,2,3,4]
输出:[2,1,4,3]
代码:
# [24] 两两交换链表中的节点
#
# @lc code=start
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
# head0 = ListNode(0,head)
if head==None or head.next==None:
return head
head0=head
headr=head.next
while head0!=None:
head1=head0
head2=head0.next
head3=head2.next
#print(head3.val)
head2.next=head1
if head3==None:
head1.next=None
#print("is 1")
return headr
if head3.next==None:
head1.next=head3
#print("is 2")
return headr
else:
head1.next=head3.next
#print(head3.val)
#print("is 0")
head0=head3
#print(head0.val)
return headr