24. 两两交换链表中的节点

因为说不能改变结点内部的值 感觉我这种方法不是很可靠

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
//改变结点前后的值
class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode temp = head;
        int temp1 = -1;
        int temp2 = -1;
        while(temp!=null&&temp.next!=null){
            temp1 =  temp.val;
            temp2 = temp.next.val;
            temp.val = temp2;
            temp.next.val = temp1;
            temp = temp.next.next;
        }
        return head;
    }
}

重复同一个行为采用递归

class Solution {
    public ListNode swapPairs(ListNode head) {
      if(head == null || head.next == null){
          return head;
      }
        ListNode next = head.next;
        head.next = swapPairs(next.next);
        next.next = head;
        return next;
    }
}