README_EN.md

61. Rotate List

中文文档

Description

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

Solutions

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        if head is None or head.next is None or k == 0:
            return head
        n = 0
        cur = head
        while cur:
            n += 1
            cur = cur.next
        k %= n
        if k == 0:
            return head
        p = q = head
        for i in range(k):
            q = q.next
        while q.next:
            p, q = p.next, q.next
        start = p.next
        p.next = None
        q.next = head
        return start

Java

/**
 * 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 || head.next == null) {
            return head;
        }
        int n = 0;
        ListNode cur = head;
        while (cur != null) {
            ++n;
            cur = cur.next;
        }
        k %= n;
        if (k == 0) {
            return head;
        }
        ListNode p = head, q = head;
        for (int i = 0; i < k; ++i) {
            q = q.next;
        }
        while (q.next != null) {
            p = p.next;
            q = q.next;
        }
        ListNode start = p.next;
        p.next = null;
        q.next = head;
        return start;
    }
}

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