力扣-24-两两交换链表中的节点

一说到链表题，最长考的方法就是递归和迭代

class Solution {
    /**
     * 迭代地实现链表中两两节点的交换
     * 新链表的头结点
     * temp表示当前到达的节点
     * 交换前：temp -> node1 -> node2
     * 交换后：temp -> node2 -> node1
     * 交换方式：
     * temp.next = node2
     * node1.next = node2.next
     * node2.next = node1
     */
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        // 创建哨兵节点
        ListNode sentinel = new ListNode(-1);
        
        sentinel.next = head;
        ListNode temp = sentinel;
        
        // temp后有至少两个节点时才需要交换
        while (temp.next != null && temp.next.next != null) {
            ListNode node1 = temp.next;
            ListNode node2 = temp.next.next;
            
            // 交换node1和node2
            temp.next = node2;
            node1.next = node2.next;
            node2.next = node1;

            // temp 重新赋值
            temp = node1;
        }
        
        return sentinel.next;
    }
}

/**
 * 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) {
        // 空链表和单节点链表直接返回
        if (head == null || head.next == null) return head;
        
        ListNode newHead = head.next;
        head.next = swapPairs(newHead.next);    // 递归地交换除前两个节点以外的节点
        newHead.next = head;
        return newHead;
    }
}

 

力扣-25-K个一组翻转链表

/**
 * 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 reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);   // 哨兵节点
        dummy.next = head;

        ListNode prev = dummy;
        ListNode end = dummy;

        while (end.next != null) {
            for (int i = 0; i < k && end != null; i++) end = end.next;
            if (end == null) break;
            ListNode start = prev.next;
            ListNode next = end.next;   // 保留下一段的开头

            end.next = null;    // 在此处断开
            prev.next = reverse(start);
            start.next = next;
            prev = start;
            end = prev;
        }
        return dummy.next;
    }

    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;

        while (curr != null) {
            ListNode next = curr.next;  // 当前节点的下一个节点
            curr.next = prev;
            prev = curr;
            curr = next;
        }

        return prev;
    }
}