【力扣刷题】24. 两两交换链表中的节点

输入：head = [1,2,3,4]
输出：[2,1,4,3]

输入：head = []
输出：[]

输入：head = [1]
输出：[1]

/**
 * 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;
        }
        //用stack保存每次迭代的两个节点
        Stack<ListNode> stack = new Stack<ListNode>();
        ListNode p = new ListNode(-1);
        ListNode cur = head;
        //head指向新的p节点，函数结束时返回head.next即可
        head = p;
        while(cur!=null &amp;&amp; cur.next!=null) {
            //将两个节点放入stack中
            stack.add(cur);
            stack.add(cur.next);
            //当前节点往前走两步
            cur = cur.next.next;
            //从stack中弹出两个节点，然后用p节点指向新弹出的两个节点
            p.next = stack.pop();
            p = p.next;
            p.next = stack.pop();
            p = p.next;
        }
        //注意边界条件，当链表长度是奇数时，cur就不为空
        if(cur!=null) {
            p.next = cur;
        } else {
            p.next = null;
        }
        return head.next;
    }
}