力扣00025.K 个一组翻转链表

发表于 分类于 阅读次数： 本文字数： 2.6k 阅读时长 ≈ 9 分钟
K 个一组翻转链表、递归、链表、困难

题目描述

给你链表的头节点 head ，每 k 个节点一组进行翻转，请你返回修改后的链表。

k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。

你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。

示例 1：

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

示例 2：

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

提示：

  • 链表中的节点数目为 n
  • 1 <= k <= n <= 5000
  • 0 <= Node.val <= 1000

进阶：你可以设计一个只用 O(1) 额外内存空间的算法解决此问题吗？

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* reverseKGroup(ListNode* head, int k) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* prev_group_end = dummy;
        while (true) {
            ListNode* group_start = prev_group_end->next;
            ListNode* group_end = getGroupEnd(group_start, k);
            if (!group_end) {
                break;
            }
            ListNode* next_group_start = group_end->next;
            group_end->next = nullptr;
            prev_group_end->next = reverseList(group_start);
            group_start->next = next_group_start;
            prev_group_end = group_start;
        }

        return dummy->next;
    }
private:
    ListNode* getGroupEnd(ListNode* start, int k) {
        for (int i = 1; i < k && start; ++i) {
            start = start->next;
        }
        return start;
    }
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = nullptr;
        ListNode* curr = head;
        while (curr) {
            ListNode* next_node = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next_node;
        }
        return prev;
    }
};

结果

执行用时 : 14 ms, 击败 29.79% 使用 C++ 的用户

内存消耗 : 14.58 MB, 击败 7.15% 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 reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prevGroupEnd = dummy;
        while (true) {
            ListNode groupStart = prevGroupEnd.next;
            ListNode groupEnd = getGroupEnd(groupStart, k);
            if (groupEnd == null) {
                break;
            }
            ListNode nextGroupStart = groupEnd.next;
            groupEnd.next = null;
            prevGroupEnd.next = reverseList(groupStart);
            groupStart.next = nextGroupStart;
            prevGroupEnd = groupStart;
        }
        return dummy.next;
    }
    private ListNode getGroupEnd(ListNode start, int k) {
        for (int i = 1; i < k && start != null; ++i) {
            start = start.next;
        }
        return start;
    }
    private ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode nextNode = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextNode;
        }
        return prev;
    }
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 Java 的用户

内存消耗 : 43.32 MB, 击败 5.99% 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 reverseKGroup(self, head, k):
        dummy = ListNode(0)
        dummy.next = head
        prev_group_end = dummy
        while True:
            group_start = prev_group_end.next
            group_end = self.get_group_end(group_start, k)
            if not group_end:
                break
            next_group_start = group_end.next
            group_end.next = None
            prev_group_end.next = self.reverse_list(group_start)
            group_start.next = next_group_start
            prev_group_end = group_start
        return dummy.next
    def get_group_end(self, start, k):
        for i in range(1, k):
            if start:
                start = start.next
        return start
    def reverse_list(self, head):
        prev = None
        curr = head
        while curr:
            next_node = curr.next
            curr.next = prev
            prev, curr = curr, next_node
        return prev

结果

执行用时 : 32 ms, 击败 67.63% 使用 Python 的用户

内存消耗 : 13.03 MB, 击败 97.28% 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 reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        dummy = ListNode(0)
        dummy.next = head
        prev_group_end = dummy
        while True:
            group_start = prev_group_end.next
            group_end = self.get_group_end(group_start, k)
            if not group_end:
                break
            next_group_start = group_end.next
            group_end.next = None
            prev_group_end.next = self.reverse_list(group_start)
            group_start.next = next_group_start
            prev_group_end = group_start
        return dummy.next
    def get_group_end(self, start, k):
        for i in range(1, k):
            if start:
                start = start.next
        return start
    def reverse_list(self, head):
        prev = None
        curr = head
        while curr:
            next_node = curr.next
            curr.next = prev
            prev, curr = curr, next_node
        return prev

结果

执行用时 : 44 ms, 击败 87.58% 使用 Python3 的用户

内存消耗 : 17.23 MB, 击败 40.79% 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* getGroupEnd(struct ListNode* start, int k);
struct ListNode* reverseList(struct ListNode* head);
struct ListNode* reverseKGroup(struct ListNode* head, int k) {
    struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode));
    dummy->next = head;
    struct ListNode* prev_group_end = dummy;
    while (1) {
        struct ListNode* group_start = prev_group_end->next;
        struct ListNode* group_end = getGroupEnd(group_start, k);
        if (!group_end) {
            break;
        }
        struct ListNode* next_group_start = group_end->next;
        group_end->next = NULL;
        prev_group_end->next = reverseList(group_start);
        group_start->next = next_group_start;
        prev_group_end = group_start;
    }
    return dummy->next;
}
struct ListNode* getGroupEnd(struct ListNode* start, int k) {
    for (int i = 1; i < k && start; ++i) {
        start = start->next;
    }
    return start;
}
struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode* prev = NULL;
    struct ListNode* curr = head;
    while (curr) {
        struct ListNode* next_node = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next_node;
    }
    return prev;
}

结果

执行用时 : 5 ms, 击败 60.44% 使用 C 的用户

内存消耗 : 6.59 MB, 击败 99.47% 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 ReverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prevGroupEnd = dummy;
        while (true) {
            ListNode groupStart = prevGroupEnd.next;
            ListNode groupEnd = GetGroupEnd(groupStart, k);
            if (groupEnd == null) {
                break;
            }
            ListNode nextGroupStart = groupEnd.next;
            groupEnd.next = null;
            prevGroupEnd.next = ReverseList(groupStart);
            groupStart.next = nextGroupStart;
            prevGroupEnd = groupStart;
        }
        return dummy.next;
    }
    private ListNode GetGroupEnd(ListNode start, int k) {
        for (int i = 1; i < k && start != null; ++i) {
            start = start.next;
        }
        return start;
    }
    private ListNode ReverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode nextNode = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextNode;
        }
        return prev;
    }
}

结果

执行用时 : 81 ms, 击败 50.00% 使用 C# 的用户

内存消耗 : 43.74 MB, 击败 5.55% 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 reverseKGroup = function(head, k) {
    let dummy = new ListNode(0);
    dummy.next = head;
    let prevGroupEnd = dummy;
    while (true) {
        let groupStart = prevGroupEnd.next;
        let groupEnd = getGroupEnd(groupStart, k);
        if (!groupEnd) {
            break;
        }
        let nextGroupStart = groupEnd.next;
        groupEnd.next = null;
        prevGroupEnd.next = reverseList(groupStart);
        groupStart.next = nextGroupStart;
        prevGroupEnd = groupStart;
    }
    return dummy.next;
    function getGroupEnd(start, k) {
        for (let i = 1; i < k && start; ++i) {
            start = start.next;
        }
        return start;
    }
    function reverseList(head) {
        let prev = null;
        let curr = head;
        while (curr) {
            let nextNode = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextNode;
        }
        return prev;
    }
};

结果

执行用时 : 84 ms, 击败 34.35% 使用 JavaScript 的用户

内存消耗 : 52.97 MB, 击败 10.36% 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 reverseKGroup(head: ListNode | null, k: number): ListNode | null {
    let dummy: ListNode = new ListNode(0);
    dummy.next = head;
    let prevGroupEnd: ListNode = dummy;
    while (true) {
        let groupStart: ListNode | null = prevGroupEnd.next;
        let groupEnd: ListNode | null = getGroupEnd(groupStart, k);
        if (!groupEnd) {
            break;
        }
        let nextGroupStart: ListNode | null = groupEnd.next;
        groupEnd.next = null;
        prevGroupEnd.next = reverseList(groupStart);
        groupStart.next = nextGroupStart;
        prevGroupEnd = groupStart;
    }
    return dummy.next;
    function getGroupEnd(start: ListNode | null, k: number): ListNode | null {
        for (let i = 1; i < k && start; ++i) {
            start = start.next;
        }
        return start;
    }
    function reverseList(head: ListNode | null): ListNode | null {
        let prev: ListNode | null = null;
        let curr: ListNode | null = head;
        while (curr) {
            let nextNode: ListNode | null = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextNode;
        }
        return prev;
    }
}

结果

执行用时 : 79 ms, 击败 79.55% 使用 TypeScript 的用户

内存消耗 : 54.74 MB, 击败 10.46% 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 int $k
     * @return ListNode
     */
    function reverseKGroup($head, $k) {
        $dummy = new ListNode(0);
        $dummy->next = $head;
        $prevGroupEnd = $dummy;
        while (true) {
            $groupStart = $prevGroupEnd->next;
            $groupEnd = $this->getGroupEnd($groupStart, $k);
            if (!$groupEnd) {
                break;
            }
            $nextGroupStart = $groupEnd->next;
            $groupEnd->next = null;
            $prevGroupEnd->next = $this->reverseList($groupStart);
            $groupStart->next = $nextGroupStart;
            $prevGroupEnd = $groupStart;
        }
        return $dummy->next;
    }
    private function getGroupEnd($start, $k) {
        for ($i = 1; $i < $k && $start; ++$i) {
            $start = $start->next;
        }
        return $start;
    }
    private function reverseList($head) {
        $prev = null;
        $curr = $head;
        while ($curr) {
            $nextNode = $curr->next;
            $curr->next = $prev;
            $prev = $curr;
            $curr = $nextNode;
        }
        return $prev;
    }
}

结果

执行用时 : 13 ms, 击败 35.71% 使用 PHP 的用户

内存消耗 : 21.41 MB, 击败 7.14% 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 reverseKGroup(_ head: ListNode?, _ k: Int) -> ListNode? {
        let dummy = ListNode(0)
        dummy.next = head
        var prevGroupEnd: ListNode? = dummy
        while true {
            let groupStart: ListNode? = prevGroupEnd?.next
            let groupEnd: ListNode? = getGroupEnd(groupStart, k)
            if groupEnd == nil {
                break
            }
            let nextGroupStart: ListNode? = groupEnd?.next
            groupEnd?.next = nil
            prevGroupEnd?.next = reverseList(groupStart)
            groupStart?.next = nextGroupStart
            prevGroupEnd = groupStart
        }
        return dummy.next
    }
    private func getGroupEnd(_ start: ListNode?, _ k: Int) -> ListNode? {
        var start = start
        for _ in 1..<k where start != nil {
            start = start?.next
        }
        return start
    }
    private func reverseList(_ head: ListNode?) -> ListNode? {
        var prev: ListNode? = nil
        var curr: ListNode? = head
        while curr != nil {
            let nextNode: ListNode? = curr?.next
            curr?.next = prev
            prev = curr
            curr = nextNode
        }
        return prev
    }
}

结果

执行用时 : 36 ms, 击败 18.75% 使用 Swift 的用户

内存消耗 : 15.75 MB, 击败 6.25% 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 reverseKGroup(head: ListNode?, k: Int): ListNode? {
        val dummy = ListNode(0)
        dummy.next = head
        var prevGroupEnd: ListNode? = dummy
        while (true) {
            val groupStart: ListNode? = prevGroupEnd?.next
            val groupEnd: ListNode? = getGroupEnd(groupStart, k)
            if (groupEnd == null) {
                break
            }
            val nextGroupStart: ListNode? = groupEnd?.next
            groupEnd?.next = null
            prevGroupEnd?.next = reverseList(groupStart)
            groupStart?.next = nextGroupStart
            prevGroupEnd = groupStart
        }
        return dummy.next
    }
    private fun getGroupEnd(start: ListNode?, k: Int): ListNode? {
        var start = start
        for (i in 1 until k) {
            if (start == null) {
                break
            }
            start = start.next
        }
        return start
    }
    private fun reverseList(head: ListNode?): ListNode? {
        var prev: ListNode? = null
        var curr: ListNode? = head
        while (curr != null) {
            val nextNode: ListNode? = curr?.next
            curr?.next = prev
            prev = curr
            curr = nextNode
        }
        return prev
    }
}

结果

执行用时 : 177 ms, 击败 87.88% 使用 Kotlin 的用户

内存消耗 : 36.66 MB, 击败 90.91% 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? reverseKGroup(ListNode? head, int k) {
    ListNode dummy = ListNode(0);
    dummy.next = head;
    ListNode? prevGroupEnd = dummy;
    while (true) {
      ListNode? groupStart = prevGroupEnd?.next;
      ListNode? groupEnd = getGroupEnd(groupStart, k);
      if (groupEnd == null) {
        break;
      }
      ListNode? nextGroupStart = groupEnd.next;
      groupEnd.next = null;
      prevGroupEnd?.next = reverseList(groupStart);
      groupStart?.next = nextGroupStart;
      prevGroupEnd = groupStart;
    }
    return dummy.next;
  }
  ListNode? getGroupEnd(ListNode? start, int k) {
    for (int i = 1; i < k && start != null; ++i) {
      start = start.next;
    }
    return start;
  }
  ListNode? reverseList(ListNode? head) {
    ListNode? prev = null;
    ListNode? curr = head;
    while (curr != null) {
      ListNode? nextNode = curr.next;
      curr.next = prev;
      prev = curr;
      curr = nextNode;
    }
    return prev;
  }
}

结果

执行用时 : 287 ms, 击败 33.33% 使用 Dart 的用户

内存消耗 : 148.09 MB, 击败 66.67% 使用 Dart 的用户

Go

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseKGroup(head *ListNode, k int) *ListNode {
    dummy := &ListNode{0, head}
    prevGroupEnd := dummy
    for {
        groupStart := prevGroupEnd.Next
        groupEnd := getGroupEnd(groupStart, k)
        if groupEnd == nil {
            break
        }
        nextGroupStart := groupEnd.Next
        groupEnd.Next = nil
        prevGroupEnd.Next = reverseList(groupStart)
        groupStart.Next = nextGroupStart
        prevGroupEnd = groupStart
    }
    return dummy.Next
}
func getGroupEnd(start *ListNode, k int) *ListNode {
    for i := 1; i < k && start != nil; i++ {
        start = start.Next
    }
    return start
}
func reverseList(head *ListNode) *ListNode {
    var prev, curr *ListNode = nil, head
    for curr != nil {
        nextNode := curr.Next
        curr.Next = prev
        prev, curr = curr, nextNode
    }
    return prev
}

结果

执行用时 : 4 ms, 击败 72.33% 使用 Go 的用户

内存消耗 : 3.39 MB, 击败 60.50% 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 reverse_k_group(head, k)
  dummy = ListNode.new(0)
  dummy.next = head
  prev_group_end = dummy
  while true
    group_start = prev_group_end.next
    group_end = get_group_end(group_start, k)
    break unless group_end
    next_group_start = group_end.next
    group_end.next = nil
    prev_group_end.next = reverse_list(group_start)
    group_start.next = next_group_start
    prev_group_end = group_start
  end
  dummy.next
end
def get_group_end(start, k)
  i = 1
  while i < k && start
    start = start.next
    i += 1
  end
  start
end
def reverse_list(head)
  prev = nil
  curr = head
  while curr
    next_node = curr.next
    curr.next = prev
    prev = curr
    curr = next_node
  end
  prev
end

结果

执行用时 : 53 ms, 击败 100.00% 使用 Ruby 的用户

内存消耗 : 206.84 MB, 击败 -% 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 reverseKGroup(head: ListNode, k: Int): ListNode = {
    var current = head
    var count = 0
    while (current != null && count < k) {
      current = current.next
      count += 1
    }
    if (count == k) {
      var prev: ListNode = null
      var nextGroupStart = current
      current = head
      for (_ <- 0 until k) {
        val nextNode = current.next
        current.next = prev
        prev = current
        current = nextNode
      }
      head.next = reverseKGroup(nextGroupStart, k)
      prev
    } else {
      head
    }
  }
}

结果

执行用时 : 565 ms, 击败 20.00% 使用 Scala 的用户

内存消耗 : 57.17 MB, 击败 20.00% 使用 Scala 的用户

Rust

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结果

执行用时 : ms, 击败 % 使用 Rust 的用户

内存消耗 : MB, 击败 % 使用 Rust 的用户

Racket

暂时未解决

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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

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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

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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