LeetCode61.旋转链表

本题有两种做法：迭代和递归
本题的本质是：将链表中后k个结点变为前k个，然后将头结点连接到尾节点

迭代

考察知识：

• 边界条件判断
• 链表倒k结点寻找
• Get思想：结环

/*** 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 rotateRight(ListNode head, int k) {if (head == null) {return null;}ListNode p = head;int count = 1;while (p.next != null) {count++;p = p.next;}k = k % count;// 成环，寻找 k 结点p.next = head;for (int i = 0; i < count - k; i++) {p = p.next;}head = p.next;p.next = null;return head;}
}

递归

暂时没思考