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。

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

解法

将链表右半部分的 k 的节点拼接到 head 即可。

注：k 对链表长度 n 取余，即 k %= n。

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 k == 0 or head is None or head.next is None:
            return head
        n, cur = 0, head
        while cur:
            n, cur = n + 1, 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 (k == 0 || head == null || head.next == null) {
            return head;
        }
        ListNode cur = head;
        int n = 0;
        while (cur != null) {
            cur = cur.next;
            ++n;
        }
        k %= n;
        if (k == 0) {
            return head;
        }
        ListNode p = head, q = head;
        while (k-- > 0) {
            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;
    }
}

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode RotateRight(ListNode head, int k) {
        if (k == 0 || head == null || head.next == null)
        {
            return head;
        }
        ListNode cur = head;
        var n = 0;
        while (cur != null)
        {
            cur = cur.next;
            ++n;
        }
        k %= n;
        if (k == 0)
        {
            return head;
        }
        ListNode p = head, q = head;
        while (k-- > 0)
        {
            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;

    }
}

Provide feedback

Saved searches

Use saved searches to filter your results more quickly

Uh oh!

Uh oh!

pFad - (p)hone/(F)rame/(a)nonymizer/(d)eclutterfier! Saves Data!