力扣00024.两两交换链表中的节点

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两两交换链表中的节点、递归、链表、中等

题目描述

给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。

示例 1：

输入：head = [1,2,3,4]
输出：[2,1,4,3]

示例 2：

输入：head = []
输出：[]

示例 3：

输入：head = [1]
输出：[1]

提示：

  • 链表中节点的数目在范围 [0, 100] 内
  • 0 <= Node.val <= 100

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* swapPairs(ListNode* head) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* current = dummy;
        while (current->next && current->next->next) {
            ListNode* node1 = current->next;
            ListNode* node2 = current->next->next;
            current->next = node2;
            node1->next = node2->next;
            node2->next = node1;
            current = node1;
        }
        return dummy->next;
    }
};

结果

执行用时 : 0 ms, 击败 100.00% 使用 C++ 的用户

内存消耗 : 9.32 MB, 击败 5.03% 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; }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode current = dummy;
        while (current.next != null && current.next.next != null) {
            ListNode node1 = current.next;
            ListNode node2 = current.next.next;
            current.next = node2;
            node1.next = node2.next;
            node2.next = node1;
            current = node1;
        }
        return dummy.next;
    }
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 Java 的用户

内存消耗 : 40.38 MB, 击败 5.18% 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 swapPairs(self, head):
        dummy = ListNode(0)
        dummy.next = head
        current = dummy
        while current.next and current.next.next:
            node1 = current.next
            node2 = current.next.next
            current.next = node2
            node1.next = node2.next
            node2.next = node1
            current = node1
        return dummy.next

结果

执行用时 : 14 ms, 击败 75.45% 使用 Python 的用户

内存消耗 : 11.44 MB, 击败 97.48% 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 swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(0)
        dummy.next = head
        current = dummy
        while current.next and current.next.next:
            node1 = current.next
            node2 = current.next.next
            current.next = node2
            node1.next = node2.next
            node2.next = node1
            current = node1
        return dummy.next

结果

执行用时 : 33 ms, 击败 90.27% 使用 Python3 的用户

内存消耗 : 16.45 MB, 击败 30.15% 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* swapPairs(struct ListNode* head) {
    struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode));
    dummy->next = head;
    struct ListNode* current = dummy;
    while (current->next && current->next->next) {
        struct ListNode* node1 = current->next;
        struct ListNode* node2 = current->next->next;
        current->next = node2;
        node1->next = node2->next;
        node2->next = node1;
        current = node1;
    }
    return dummy->next;
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 C 的用户

内存消耗 : 5.71 MB, 击败 95.91% 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 SwapPairs(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode current = dummy;
        while (current.next != null && current.next.next != null) {
            ListNode node1 = current.next;
            ListNode node2 = current.next.next;
            current.next = node2;
            node1.next = node2.next;
            node2.next = node1;
            current = node1;
        }
        return dummy.next;
    }
}

结果

执行用时 : 57 ms, 击败 88.57% 使用 C# 的用户

内存消耗 : 39.66 MB, 击败 5.71% 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
 * @return {ListNode}
 */
var swapPairs = function(head) {
    let dummy = new ListNode(0);
    dummy.next = head;
    let current = dummy;
    while (current.next && current.next.next) {
        let node1 = current.next;
        let node2 = current.next.next;
        current.next = node2;
        node1.next = node2.next;
        node2.next = node1;
        current = node1;
    }
    return dummy.next;
};

结果

执行用时 : 37 ms, 击败 100.00% 使用 JavaScript 的用户

内存消耗 : 49.33 MB, 击败 5.29% 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 swapPairs(head: ListNode | null): ListNode | null {
    let dummy: ListNode = new ListNode(0);
    dummy.next = head;
    let current: ListNode | null = dummy;
    while (current?.next && current?.next?.next) {
        let node1: ListNode | null = current.next;
        let node2: ListNode | null = current.next.next;
        if (node1 && node2) {
            current.next = node2;
            node1.next = node2.next;
            node2.next = node1;
            current = node1;
        }
    }
    return dummy.next;
}

结果

执行用时 : 60 ms, 击败 86.10% 使用 TypeScript 的用户

内存消耗 : 51.79 MB, 击败 5.02% 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 $head
     * @return ListNode
     */
    function swapPairs($head) {
        $dummy = new ListNode(0);
        $dummy->next = $head;
        $current = $dummy;
        while ($current->next && $current->next->next) {
            $node1 = $current->next;
            $node2 = $current->next->next;
            $current->next = $node2;
            $node1->next = $node2->next;
            $node2->next = $node1;
            $current = $node1;
        }
        return $dummy->next;
    }
}

结果

执行用时 : 11 ms, 击败 16.67% 使用 PHP 的用户

内存消耗 : 19.93 MB, 击败 5.55% 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 swapPairs(_ head: ListNode?) -> ListNode? {
        let dummy = ListNode(0)
        dummy.next = head
        var current: ListNode? = dummy
        while current?.next != nil && current?.next?.next != nil {
            let node1 = current?.next
            let node2 = current?.next?.next
            current?.next = node2
            node1?.next = node2?.next
            node2?.next = node1
            current = node1
        }
        return dummy.next
    }
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 Swift 的用户

内存消耗 : 15.22 MB, 击败 20.67% 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 swapPairs(head: ListNode?): ListNode? {
        val dummy = ListNode(0)
        dummy.next = head
        var current: ListNode? = dummy
        while (current?.next != null && current.next?.next != null) {
            val node1 = current.next
            val node2 = current.next?.next
            current.next = node2
            node1?.next = node2?.next
            node2?.next = node1
            current = node1
        }
        return dummy.next
    }
}

结果

执行用时 : 138 ms, 击败 81.08% 使用 Kotlin 的用户

内存消耗 : 33.44 MB, 击败 32.43% 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? swapPairs(ListNode? head) {
    ListNode? tail = head;
    for (int i = 0; i < 2; i++) {
      if (tail == null) {
        return head;
      }
      tail = tail.next;
    }
    ListNode? pre;
    ListNode? cur = head;
    while (cur != tail) {
      ListNode? temp = cur?.next;
      cur?.next = pre;
      pre = cur;
      cur = temp;
    }
    head?.next = swapPairs(tail);
    return pre;
  }
}

结果

执行用时 : 302 ms, 击败 22.22% 使用 Dart 的用户

内存消耗 : 147.56 MB, 击败 55.56% 使用 Dart 的用户

Go

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func swapPairs(head *ListNode) *ListNode {
	dummy := &ListNode{Val: 0, Next: head}
	current := dummy
	for current.Next != nil && current.Next.Next != nil {
		node1 := current.Next
		node2 := current.Next.Next
		current.Next = node2
		node1.Next = node2.Next
		node2.Next = node1
		current = node1
	}
	return dummy.Next
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 Go 的用户

内存消耗 : 1.97 MB, 击败 26.06% 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
# @return {ListNode}
def swap_pairs(head)
  dummy = ListNode.new(0)
  dummy.next = head
  current = dummy
  while current.next && current.next.next
    node1 = current.next
    node2 = current.next.next
    current.next = node2
    node1.next = node2.next
    node2.next = node1
    current = node1
  end
  dummy.next
end

结果

执行用时 : 47 ms, 击败 100.00% 使用 Ruby 的用户

内存消耗 : 206.64 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 swapPairs(head: ListNode): ListNode = {
    if (head == null || head.next == null) {
      return head
    }
    val t1 = head.next
    val t2 = head.next.next
    t1.next = head
    head.next = swapPairs(t2)
    t1
  }
}

结果

执行用时 : 460 ms, 击败 85.71% 使用 Scala 的用户

内存消耗 : 56.41 MB, 击败 14.29% 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 swap_pairs(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        if let Some(mut node) = head {
            if let Some(mut next) = node.next.take() {
                node.next = Self::swap_pairs(next.next.take());
                next.next = Some(node);
                Some(next)
            } else {
                Some(node)
            }
        } else {
            None
        }
    }
}

结果

执行用时 : 0 ms, 击败 100.00% 使用 Rust 的用户

内存消耗 : 2.15 MB, 击败 21.54% 使用 Rust 的用户

Racket

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; Definition for singly-linked list:
#|

; val : integer?
; next : (or/c list-node? #f)
(struct list-node
  (val next) #:mutable #:transparent)

; constructor
(define (make-list-node [val 0])
  (list-node val #f))

|#

(define (swap-pairs head)
  (cond
    ((or (not head) (not (list-node? head)) (not (list-node? (list-node-next head))))
     head)
    (else
     (let ((node1 head)
           (node2 (list-node-next head))
           (rest (list-node-next (list-node-next head))))
       (set-list-node-next! node1 (swap-pairs rest))
       (set-list-node-next! node2 node1)
       node2))))

结果

执行用时 : 175 ms, 击败 100.00% 使用 Racket 的用户

内存消耗 : 98.80 MB, 击败 100.00% 使用 Racket 的用户

Erlang

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%% Definition for singly-linked list.
%%
%% -record(list_node, {val = 0 :: integer(),
%%                     next = null :: 'null' | #list_node{}}).

-spec swap_pairs(Head :: #list_node{} | null) -> #list_node{} | null.
swap_pairs(Head) when Head =:= null -> null;
swap_pairs(Head) when is_record(Head, list_node) -> 
    case Head#list_node.next of
        null -> Head;
        Next -> 
            NewNext = swap_pairs(Next#list_node.next),
            NewHead = Head#list_node{next = NewNext},
            Next#list_node{next = NewHead}
    end.

结果

执行用时 : 222 ms, 击败 -% 使用 Erlang 的用户

内存消耗 : 59.48 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