java - 链表中的交换节点溢出Java

/**
 * 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 curr = head;//current node in list we are looking at
        ListNode retList = new ListNode(0);//new linked list to return
        ListNode newHead = retList;//reference to head of the list
        while(curr.next != null); {//while there is another point in the original list
            retList.val = curr.next.val;//the value in the return list = next value in original lis
            retList.next = new ListNode(curr.val);//opposite of last line+creates another node in return list
            if(curr.next.next == null) {//if there is another node after the next one
                curr = curr.next;//if there isnt then the while loop wont go
            }
            else {
                retList.next.next = new ListNode(0);//creates a new node 2 nodes ahead
                retList = retList.next.next;//goes to that node
                curr = curr.next.next;//jumps 2 nodes in the original list
            }
        }
        return newHead;//returns head to the list we are returning
    }
}

我真的不知道为什么这不起作用，while循环应该结束。提示是将列表中的每个节点与列表中的每个下一个节点交换
1,2,3,4变成2,1,4,3。它在leetcode https://leetcode.com/problems/swap-nodes-in-pairs/description/上
每次我运行代码时，都会给我一个超时错误。

最佳答案

您的解决方案是正确的，但是您的时间已经过去了。

关于java - 链表中的交换节点溢出Java，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/49838715/