力扣00061.旋转链表

发表于 分类于 阅读次数： 本文字数： 2.3k 阅读时长 ≈ 8 分钟
旋转链表、链表、双指针、中等

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
  • $0 <= k <= 2 * 10^9$

解决方法

C++

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head || k == 0) {
            return head;
        }
        int length = 1;
        ListNode *current = head;
        while (current->next) {
            length++;
            current = current->next;
        }
        k = k % length;
        if (k == 0) {
            return head;
        }
        ListNode *fast = head;
        ListNode *slow = head;
        for (int i = 0; i < k; i++) {
            fast = fast->next;
        }
        while (fast->next) {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode *new_head = slow->next;
        slow->next = NULL;
        fast->next = head;
        return new_head;
    }
};

结果

执行用时 : 5 ms, 击败 53.76% 使用 C++ 的用户

内存消耗 : 14.91 MB, 击败 13.56% 使用 C++ 的用户

Java

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/**
 * 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 || k == 0) {
            return head;
        }
        int length = 1;
        ListNode current = head;
        while (current.next != null) {
            length++;
            current = current.next;
        }
        k = k % length;
        if (k == 0) {
            return head;
        }
        ListNode fast = head;
        ListNode slow = head;
        for (int i = 0; i < k; i++) {
            fast = fast.next;
        }
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        ListNode newHead = slow.next;
        slow.next = null;
        fast.next = head;
        return newHead;
    }
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 Java 的用户

内存消耗 : 41.34 MB, 击败 52.48% 使用 Java 的用户

Python

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head or k == 0:
            return head
        length = 1
        current = head
        while current.next:
            length += 1
            current = current.next
        k = k % length
        if k == 0:
            return head
        fast = head
        slow = head
        for _ in range(k):
            fast = fast.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        new_head = slow.next
        slow.next = None
        fast.next = head
        return new_head

结果

执行用时 : 23 ms, 击败 42.32% 使用 Python 的用户

内存消耗 : 11.53 MB, 击败 71.87% 使用 Python 的用户

Python3

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# 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: Optional[ListNode], k: int) -> Optional[ListNode]:
        if not head or k == 0:
            return head
        length = 1
        current = head
        while current.next:
            length += 1
            current = current.next
        k = k % length
        if k == 0:
            return head
        fast = head
        slow = head
        for _ in range(k):
            fast = fast.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        new_head = slow.next
        slow.next = None
        fast.next = head
        return new_head

结果

执行用时 : 46 ms, 击败 29.19% 使用 Python3 的用户

内存消耗 : 16.41 MB, 击败 57.60% 使用 Python3 的用户

C

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* rotateRight(struct ListNode* head, int k) {
    if (!head || k == 0) {
        return head;
    }
    int length = 1;
    struct ListNode *current = head;
    while (current->next) {
        length++;
        current = current->next;
    }
    k = k % length;
    if (k == 0) {
        return head;
    }
    struct ListNode *fast = head;
    struct ListNode *slow = head;
    for (int i = 0; i < k; i++) {
        fast = fast->next;
    }
    while (fast->next) {
        fast = fast->next;
        slow = slow->next;
    }
    struct ListNode *newHead = slow->next;
    slow->next = NULL;
    fast->next = head;
    return newHead;
}

结果

执行用时 : 2 ms, 击败 74.04% 使用 C 的用户

内存消耗 : 5.82 MB, 击败 91.17% 使用 C 的用户

C#

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/**
 * 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 (head == null || k == 0) {
            return head;
        }
        int length = 1;
        ListNode current = head;
        while (current.next != null) {
            length++;
            current = current.next;
        }
        k = k % length;
        if (k == 0) {
            return head;
        }
        ListNode fast = head;
        ListNode slow = head;
        for (int i = 0; i < k; i++) {
            fast = fast.next;
        }
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        ListNode newHead = slow.next;
        slow.next = null;
        fast.next = head;
        return newHead;
    }
}

结果

执行用时 : 64 ms, 击败 72.18% 使用 C# 的用户

内存消耗 : 41.42 MB, 击败 27.07% 使用 C# 的用户

JavaScript

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/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
var rotateRight = function(head, k) {
    if (!head || k === 0) {
        return head;
    }
    let length = 1;
    let current = head;
    while (current.next) {
        length++;
        current = current.next;
    }
    k = k % length;
    if (k === 0) {
        return head;
    }
    let fast = head;
    let slow = head;
    for (let i = 0; i < k; i++) {
        fast = fast.next;
    }
    while (fast.next) {
        fast = fast.next;
        slow = slow.next;
    }
    let newHead = slow.next;
    slow.next = null;
    fast.next = head;
    return newHead;
};

结果

执行用时 : 64 ms, 击败 81.59% 使用 JavaScript 的用户

内存消耗 : 51.45 MB, 击败 13.45% 使用 JavaScript 的用户

TypeScript

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/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function rotateRight(head: ListNode | null, k: number): ListNode | null {
    if (!head || k === 0) {
        return head;
    }
    let length = 1;
    let current = head;
    while (current.next) {
        length++;
        current = current.next;
    }
    k = k % length;
    if (k === 0) {
        return head;
    }
    let fast: ListNode | null = head;
    let slow: ListNode | null = head;
    for (let i = 0; i < k; i++) {
        if (fast) {
            fast = fast.next;
        }
    }
    while (fast && fast.next) {
        fast = fast.next;
        slow = slow?.next || null;
    }
    let newHead: ListNode | null = slow?.next || null;
    if (slow) {
        slow.next = null;
    }
    if (fast) {
        fast.next = head;
    }
    return newHead;
}

结果

执行用时 : 68 ms, 击败 80.00% 使用 TypeScript 的用户

内存消耗 : 52.75 MB, 击败 5.56% 使用 TypeScript 的用户

PHP

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/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val = 0, $next = null) {
 *         $this->val = $val;
 *         $this->next = $next;
 *     }
 * }
 */
class Solution {

    /**
     * @param ListNode $head
     * @param Integer $k
     * @return ListNode
     */
    function rotateRight($head, $k) {
        if (!$head || $k === 0) {
            return $head;
        }
        $length = 1;
        $current = $head;
        while ($current->next) {
            $length++;
            $current = $current->next;
        }
        $k = $k % $length;
        if ($k === 0) {
            return $head;
        }
        $fast = $head;
        $slow = $head;
        for ($i = 0; $i < $k; $i++) {
            $fast = $fast->next;
        }
        while ($fast->next) {
            $fast = $fast->next;
            $slow = $slow->next;
        }
        $newHead = $slow->next;
        $slow->next = null;
        $fast->next = $head;
        return $newHead;
    }
}

结果

执行用时 : 12 ms, 击败 50.00% 使用 PHP 的用户

内存消耗 : 20.07 MB, 击败 10.00% 使用 PHP 的用户

Swift

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func rotateRight(_ head: ListNode?, _ k: Int) -> ListNode? {
        guard let head = head, k > 0 else {
            return head
        }
        var length = 1
        var current = head
        while current.next != nil {
            length += 1
            current = current.next!
        }
        let rotateSteps = k % length
        if rotateSteps == 0 {
            return head
        }
        var fast: ListNode? = head
        var slow: ListNode? = head
        for _ in 0..<rotateSteps {
            fast = fast?.next
        }
        while fast?.next != nil {
            fast = fast?.next
            slow = slow?.next
        }
        let newHead = slow?.next
        slow?.next = nil
        fast?.next = head
        return newHead
    }
}

结果

执行用时 : 9 ms, 击败 70.89% 使用 Swift 的用户

内存消耗 : 15.43 MB, 击败 11.39% 使用 Swift 的用户

Kotlin

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/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun rotateRight(head: ListNode?, k: Int): ListNode? {
        if (head == null || k == 0) {
            return head
        }
        var length = 1
        var current = head
        while (current?.next != null) {
            length++
            current = current.next
        }
        val rotateSteps = k % length
        if (rotateSteps == 0) {
            return head
        }
        var fast: ListNode? = head
        var slow: ListNode? = head
        repeat(rotateSteps) {
            fast = fast?.next
        }
        while (fast?.next != null) {
            fast = fast?.next
            slow = slow?.next
        }
        val newHead = slow?.next
        slow?.next = null
        fast?.next = head
        return newHead
    }
}

结果

执行用时 : 167 ms, 击败 90.70% 使用 Kotlin 的用户

内存消耗 : 35.81 MB, 击败 30.23% 使用 Kotlin 的用户

Dart

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/**
 * Definition for singly-linked list.
 * class ListNode {
 *   int val;
 *   ListNode? next;
 *   ListNode([this.val = 0, this.next]);
 * }
 */
class Solution {
  ListNode? rotateRight(ListNode? head, int k) {
      if (head == null || k == 0) {
    return head;
    }
    int length = 1;
    ListNode? current = head;
    while (current?.next != null) {
        length++;
        current = current?.next;
    }
    k = k % length;
    if (k == 0) {
        return head;
    }
    ListNode? fast = head;
    ListNode? slow = head;
    for (int i = 0; i < k; i++) {
        fast = fast?.next;
    }
    while (fast?.next != null) {
        fast = fast?.next;
        slow = slow?.next;
    }
    ListNode? newHead = slow?.next;
    slow?.next = null;
    fast?.next = head;
    return newHead;
    }
}

结果

执行用时 : 292 ms, 击败 100.00% 使用 Dart 的用户

内存消耗 : 144.09 MB, 击败 100.00% 使用 Dart 的用户

Go

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func rotateRight(head *ListNode, k int) *ListNode {
    if head == nil || k == 0 {
        return head
    }
    length := 1
    current := head
    for current.Next != nil {
        length++
        current = current.Next
    }
    k = k % length
    if k == 0 {
        return head
    }
    fast, slow := head, head
    for i := 0; i < k; i++ {
        fast = fast.Next
    }
    for fast.Next != nil {
        fast = fast.Next
        slow = slow.Next
    }
    newHead := slow.Next
    slow.Next = nil
    fast.Next = head
    return newHead
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 Go 的用户

内存消耗 : 2.36 MB, 击败 34.33% 使用 Go 的用户

Ruby

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# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} head
# @param {Integer} k
# @return {ListNode}
def rotate_right(head, k)
  return head if head.nil? || k.zero?
  length = 1
  current = head
  while current.next
    length += 1
    current = current.next
  end
  k = k % length
  return head if k.zero?
  fast = head
  slow = head
  k.times { fast = fast.next }
  while fast.next
    fast = fast.next
    slow = slow.next
  end
  new_head = slow.next
  slow.next = nil
  fast.next = head
  new_head
end

结果

执行用时 : 64 ms, 击败 25.00% 使用 Ruby 的用户

内存消耗 : 206.50 MB, 击败 50.00% 使用 Ruby 的用户

Scala

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/**
 * Definition for singly-linked list.
 * class ListNode(_x: Int = 0, _next: ListNode = null) {
 *   var next: ListNode = _next
 *   var x: Int = _x
 * }
 */
object Solution {
  def rotateRight(head: ListNode, k: Int): ListNode = {
    if (head == null || k == 0) {
      return head
    }
    var length = 1
    var current = head
    while (current.next != null) {
      length += 1
      current = current.next
    }
    val rotateSteps = k % length
    if (rotateSteps == 0) {
      return head
    }
    var fast: ListNode = head
    var slow: ListNode = head
    for (_ <- 0 until rotateSteps) {
      fast = fast.next
    }
    while (fast.next != null) {
      fast = fast.next
      slow = slow.next
    }
    val newHead: ListNode = slow.next
    slow.next = null
    fast.next = head
    newHead
  }
}

结果

执行用时 : 513 ms, 击败 50.00% 使用 Scala 的用户

内存消耗 : 56.80 MB, 击败 100.00% 使用 Scala 的用户

Rust

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// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn rotate_right(mut head: Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {
        let mut len = 0;
        let mut p = &head;
        while let Some(node) = p {
            len += 1;
            p = &node.next;
        }
        if len <= 1 {
            return head;
        }
        let k = k % len;
        if k == 0 {
            return head;
        }
        let mut p = Self::get_last_nth(&mut head, k);
        let mut q = &mut p;
        while (*q).is_some() && q.as_ref().unwrap().next.is_some() {
            q = &mut q.as_mut().unwrap().next;
        }
        q.as_mut().unwrap().next = head;
        p
    }
    fn get_last_nth(head: &mut Option<Box<ListNode>>, n: i32) -> Option<Box<ListNode>> {
        unsafe {
            let mut slow = head as *mut Option<Box<ListNode>>;
            let mut fast = head as *mut Option<Box<ListNode>>;
            for _ in 0..n {
                fast = &mut (*fast).as_mut().unwrap().next;
            }
            while (*fast).is_some() {
                slow = &mut (*slow).as_mut().unwrap().next;
                fast = &mut (*fast).as_mut().unwrap().next;
            }
            (*slow).take()
        }
    }
}

结果

执行用时 : 1 ms, 击败 24.14% 使用 Rust 的用户

内存消耗 : 2.13 MB, 击败 31.03% 使用 Rust 的用户

Racket

暂时未解决

1

结果

执行用时 : ms, 击败 % 使用 Racket 的用户

内存消耗 : MB, 击败 % 使用 Racket 的用户

Erlang

暂时未解决

1

结果

执行用时 : ms, 击败 % 使用 Erlang 的用户

内存消耗 : MB, 击败 % 使用 Erlang 的用户

Elixir

暂时未解决

1

结果

执行用时 : ms, 击败 % 使用 Elixir 的用户

内存消耗 : MB, 击败 % 使用 Elixir 的用户