链表问题总结

1. 递归反转链表
   （1）反转整个链表

/**
 * 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 reverseList(ListNode head) {
        if (head == null || head.next == null) return head;//base case:empty or only one node, reverse与否一个样，so simply return head
        ListNode last = reverse(head.next);//recursion: new head = reverse(old head)先递归再对当前node进行操作，那么会先递归到尽头，然后从后向前依次执行操作，即修改link
        head.next.next = head;//modify link: point the last node of the reversed list to head
        head.next = null;//modify link:point head to null
        return last;//return new head
    }
}

（2）反转前N个节点
（3）反转M节点到N节点
（4）K个一组反转链表