力扣刷题61.旋转列表(java)

Given a linked list, rotate the list to the right by k places, where k is non-negative.
给一个链表，旋转链表 k 次
Example 1:

Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

题解

拼接法：
移动后拼接。

连成环的解法：
设立一个头指针和尾指针，遍历一次链表得到链表的长度l。并将尾指针的next指向头指针，同时将head指针和尾指针向后移动L-k个元素，返回新head

class Solution {
    public ListNode rotateRight(ListNode head, int k) {
          ListNode loap = head;
          if(head==null) return null;
          int len = 0;
          while (loap != null){
              len++;
              loap = loap.next;
          }
          k = k % len;
        if(k == 0) return head;
          int i = len - k;
        ListNode temp1 = head;
        ListNode temp2 = head;
        while (i > 0){
              i--;
              temp2 = temp1;
              temp1 = temp1.next;
        }
        temp2.next = null;
        ListNode newHead = temp1;
        while (k -1 >0){
            k--;
            temp1 = temp1.next;
        }
        temp1.next = head;
        return newHead;
    }
}