Question

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：递归

我们可以通过递归的方式实现两两交换链表中的节点。

递归的终止条件是链表中没有节点，或者链表中只有一个节点，此时无法进行交换，直接返回该节点。

否则，我们递归交换链表 $head.next.next$，记交换后的头节点为 $t$，然后我们记 $head$ 的下一个节点为 $p$，然后令 $p$ 指向 $head$，而 $head$ 指向 $t$，最后返回 $p$。

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

# 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]:
    if head is None or head.next is None:
      return head
    t = self.swapPairs(head.next.next)
    p = head.next
    p.next = head
    head.next = t
    return p

/**
 * 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) {
    if (head == null || head.next == null) {
      return head;
    }
    ListNode t = swapPairs(head.next.next);
    ListNode p = head.next;
    p.next = head;
    head.next = t;
    return p;
  }
}

/**
 * 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) {
    if (!head || !head->next) {
      return head;
    }
    ListNode* t = swapPairs(head->next->next);
    ListNode* p = head->next;
    p->next = head;
    head->next = t;
    return p;
  }
};

/**
 * 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
  }
  t := swapPairs(head.Next.Next)
  p := head.Next
  p.Next = head
  head.Next = t
  return p
}

/**
 * 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 {
  if (!head || !head.next) {
    return head;
  }
  const t = swapPairs(head.next.next);
  const p = head.next;
  p.next = head;
  head.next = t;
  return p;
}

// 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
  }
}

/**
 * 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) {
  if (!head || !head.next) {
    return head;
  }
  const t = swapPairs(head.next.next);
  const p = head.next;
  p.next = head;
  head.next = t;
  return p;
};

# 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

方法二：迭代

我们设置一个虚拟头节点 $dummy$，初始时指向 $head$，然后设置两个指针 $pre$ 和 $cur$，初始时 $pre$ 指向 $dummy$，而 $cur$ 指向 $head$。

接下来，我们遍历链表，每次需要交换 $pre$ 后面的两个节点，因此我们先判断 $cur$ 和 $cur.next$ 是否为空，若不为空，则进行交换，否则终止循环。

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

# 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(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

/**
 * 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;
    ListNode 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;
  }
}

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

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func swapPairs(head *ListNode) *ListNode {
  dummy := &ListNode{Next: 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
}

/**
 * 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 [pre, cur] = [dummy, 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;
}

/**
 * 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, cur] = [dummy, 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;
};

# Definition for singly-linked list.
# class ListNode {
#  public $val;
#  public $next;
#  public function __construct($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;
    $prev = $dummy;

    while ($head !== null && $head->next !== null) {
      $first = $head;
      $second = $head->next;

      $first->next = $second->next;
      $second->next = $first;
      $prev->next = $second;

      $prev = $first;
      $head = $first->next;
    }

    return $dummy->next;
  }
}