Question:

Given a linked list, rotate the list to the right by k places, where k is non-negative.

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

Thinking:

  • Method:
    • 首先让链表头尾相接，然后找到新的头部，前一个元素就是其尾部。这种方法效率不高。
    • 二刷时需要更新。
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head == null) return null;
        ListNode cur = head;
        int count = 1;
        while(cur.next != null){
            count ++;
            cur = cur.next;
        }
        cur.next = head;
        int num = count - k % count;
        cur = head;
        while(--num >= 0){
            cur = cur.next;
        }
        ListNode result = cur;
        while(cur.next != result){
            cur = cur.next;
        }
        cur.next = null;
        return result;
    }
}

二刷

现在发现说话真的是不要钱，一刷的时候说要什么二刷提高效率就提高效率？

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head == null) return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;
        int len = 0;
        while(cur.next != null){
            len ++;
            cur = cur.next;
        }
        if(len == 0) return head;
        int p = len - k % len;
        if(p == len) return head;
        ListNode end = cur;
        cur = dummy;
        while(p-- > 0){
            cur = cur.next;
        }
        ListNode result = cur.next;
        cur.next = null;
        end.next = dummy.next;
        return result;
    }
}