61. 旋转链表

题目描述

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。

输入输出

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输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]

输入：head = [0,1,2], k = 4
输出：[2,0,1]

基本思路

  1. 先求出链表长度length和标记尾节点指针tail
  2. new一个新指针p 往后移动length - k%length - 1步到达新链表头结点
  3. p -> null head = p.next tail -> head

java实现

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class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head == null || k == 0) return head;
        int length = 0;
        ListNode tail = null;
        for(ListNode p = head; p != null; p = p.next) {
            tail = p;
            length++;
        }
        k = k%length;
        int move = length - k - 1;//move = 3
        ListNode p = head;
        for(int i = 0; i < move; i++){
            p = p.next;
        }
        tail.next = head;
        head = p.next;
        p.next = null;
        return head;

    }
}