61. 旋转链表

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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。

示例 1:

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

示例 2:

输入:head = [0,1,2], k = 4
输出:[2,0,1]

提示:

  • 链表中节点的数目在范围 [0, 500] 内
  • -100 <= Node.val <= 100
  • 0 <= k <= 2 * 109

/**
 * 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(k == 0 || head == null || head.next == null) {
            return head;
        }
        int n = 1;
        ListNode iter = head;
        //算长度
        while(iter.next != null) {
            iter = iter.next;
            n++;
        }
        //取余
        int add = n - k % n;
        if(add == n) {
            return head;
        }
        //连成环
        iter.next = head;
        //找到尾部位置
        while(add-- > 0 ) {
            iter = iter.next;
        }
        //指向头
        ListNode ret = iter.next;
        //断开尾->头
        iter.next = null;
        return ret;

    }
}