README.md

24. 两两交换链表中的节点

English Version

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

解法

方法一：迭代

设置虚拟头节点 dummy，pre 指针初始指向 dummy，遍历链表，每次交换 pre 后面的两个节点即可。

时间复杂度为 $O(n)$，空间复杂度为 $O(1)$，其中 $n$ 是链表的长度。

方法二：递归

时间复杂度为 $O(n)$，空间复杂度为 $O(n)$，其中 $n$ 是链表的长度。

Python3

# 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: ListNode) -> ListNode:
        dummy = ListNode(next=head)
        pre, cur = dummy, head
        while cur and cur.next:
            t = cur.next
            cur.next = t.next
            t.next = cur
            pre.next = t
            pre, cur = cur, cur.next
        return dummy.next

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 swapPairs(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy, cur = head;
        while (cur != null && cur.next != null) {
            ListNode t = cur.next;
            cur.next = t.next;
            t.next = cur;
            pre.next = t;
            pre = cur;
            cur = cur.next;
        }
        return dummy.next;
    }
}

JavaScript

/**
 * 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) {
    const dummy = new ListNode(0, head);
    let pre = dummy;
    let cur = head;
    while (cur && cur.next) {
        const t = cur.next;
        cur.next = t.next;
        t.next = cur;
        pre.next = t;
        pre = cur;
        cur = cur.next;
    }
    return dummy.next;
};

C++

/**
 * 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, head);
        ListNode *pre = dummy, *cur = head;
        while (cur != nullptr && cur->next != nullptr) {
            ListNode* t = cur->next;
            cur->next = t->next;
            t->next = cur;
            pre->next = t;
            pre = cur;
            cur = cur->next;
        }
        return dummy->next;
    }
};

Go

迭代：

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

递归：

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func swapPairs(head *ListNode) *ListNode {
	if head == nil || head.Next == nil {
		return head
	}
	res := swapPairs(head.Next.Next)
	p := head.Next
	p.Next, head.Next = head, res
	return p
}

Ruby

# 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, head)
    pre = dummy
    cur = head
    while !cur.nil? && !cur.next.nil?
        t = cur.next
        cur.next = t.next
        t.next = cur
        pre.next = t
        pre = cur
        cur = cur.next
    end
    dummy.next
end

TypeScript

/**
 * 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 {
    const dummy = new ListNode(0, head);
    let cur = dummy;
    while (cur.next != null && cur.next.next != null) {
        const a = cur.next;
        const b = cur.next.next;
        [a.next, b.next, cur.next] = [b.next, a, b];
        cur = cur.next.next;
    }
    return dummy.next;
}

Rust

// 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>> {
        let mut dummy = Some(Box::new(ListNode { val: 0, next: head }));
        let mut cur = dummy.as_mut().unwrap();
        while cur.next.is_some() && cur.next.as_ref().unwrap().next.is_some() {
            cur.next = {
                let mut b = cur.next.as_mut().unwrap().next.take();
                cur.next.as_mut().unwrap().next = b.as_mut().unwrap().next.take();
                let a = cur.next.take();
                b.as_mut().unwrap().next = a;
                b
            };
            cur = cur.next.as_mut().unwrap().next.as_mut().unwrap();
        }
        dummy.unwrap().next
    }
}

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