1、题目来源
24. 两两交换链表中的节点 - 力扣(LeetCode)
2、题目描述
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
3、题解分享
class Solution {
public ListNode swapPairs(ListNode head) {
// 思路:递归 + 一步步交换后面的节点
if (head == null || head.next == null) {
return head;
}
ListNode newHead = head.next;
head.next = swapPairs(newHead.next);
newHead.next = head;
return newHead;
}
}
/**
* 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) {
// 思路:定义哑结点 + 模拟指针切换
if(head == null || head.next == null){
return head;
}
ListNode dummy = new ListNode();
ListNode pre = dummy;
ListNode cur = head;
ListNode after = head.next;
while(cur != null && after != null){
pre.next = after;
cur.next = after.next;
after.next = cur;
pre = cur;
cur = cur.next;
if(cur != null)
after = cur.next;
}
return dummy.next;
}
}