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]
示例 2:
输入: head = []
输出: []
示例 3:
输入: head = [1]
输出: [1]
提示:
- 链表中节点的数目在范围
[0, 100]内 0 <= Node.val <= 100
解题答案
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) return head;
ListNode dummyNode = new ListNode(0, head);
ListNode pre = dummyNode;
while (pre.next != null && pre.next.next != null) {
ListNode first = pre.next;
ListNode second = first.next;
pre.next = second;
first.next = second.next;
second.next = first;
pre = first;
}
return dummyNode.next;
}
}