官方链接
https://leetcode-cn.com/problems/swap-nodes-in-pairs/
给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
示例:
给定 1->2->3->4, 你应该返回 2->1->4->3.解法一
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) return head;
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode c = dummy;
while(c.next != null && c.next.next != null) {
ListNode a = c.next;
ListNode b = c.next.next;
a.next = b.next;
c.next = b;
b.next = a;
c = c.next.next;
}
return dummy.next;
}
}
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
dummy = ListNode(-1)
dummy.next = head
c = dummy
while c.next and c.next.next:
a, b = c.next, c.next.next
a.next, c.next = b.next, b
b.next = a
c = c.next.next
return dummy.next
优化
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
pre, pre.next = self, head
while pre.next and pre.next.next:
a = pre.next
b = a.next
pre.next, a.next, b.next = b, b.next, a
pre = a
return self.next