[数据结构与算法]24.两两交换链表中的节点

24. 两两交换链表中的节点

方法一：迭代，虚拟头结点

class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode dummyNode = new ListNode(0);
        dummyNode.next = head;
        ListNode prev = dummyNode;
        while(prev.next != null && prev.next.next != null){
            ListNode temp = head.next.next; //缓存next
            prev.next = head.next;          //将prev的next改为head的next
            head.next.next = head;          //将head.next(prev.next)的next,指向head
            head.next = temp;               //将head的next接上缓存的temp
            prev = head;                    //步进1位
            head = head.next;               //步进1位
        }
        return dummyNode.next;
    }
}

方法二：1.递归

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

		two.next = one;
		one.next = swapPairs(three);

		return two;
    }
}

方法二：2.递归

class Solution {
    public ListNode swapPairs(ListNode head) {
        // base case 退出提交
        if(head == null || head.next == null){return head;}
        // 获取当前结点的下一个节点
        ListNode next = head.next;
        // 进行递归
        ListNode newNode = swapPairs(next.next);
        // 这里进行交换
        next.next = head;
        head.next = newNode;

        return next;
    }
}