思路:递归,然后交换相邻的两个值。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode swapPairs(ListNode head) {
if (head==null||head.next==null) {
return head;
}else {
int val=head.val;
head.val=head.next.val;
head.next.val=val;
head.next.next=swapPairs(head.next.next);
}
return head;
}
}
耗时:300ms,中下游