24. Swap Nodes in Pairs

Description

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)

Example 1:

Input: head = [1,2,3,4]
Output: [2,1,4,3]

Example 2:

Input: head = []
Output: []

Example 3:

Input: head = [1]
Output: [1]

Constraints:

The number of nodes in the list is in the range [0, 100].
0 <= Node.val <= 100

Solutions

Solution 1: Recursion

We can implement swapping two nodes in the linked list through recursion.

The termination condition of recursion is that there are no nodes in the linked list, or there is only one node in the linked list. At this time, swapping cannot be performed, so we directly return this node.

Otherwise, we recursively swap the linked list $head.next.next$, and let the swapped head node be $t$. Then we let $p$ be the next node of $head$, and let $p$ point to $head$, and $head$ point to $t$, finally return $p$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the linked list.

Solution 2: Iteration

We set a dummy head node $dummy$, initially pointing to $head$, and then set two pointers $pre$ and $cur$, initially $pre$ points to $dummy$, and $cur$ points to $head$.

Next, we traverse the linked list. Each time we need to swap the two nodes after $pre$, so we first judge whether $cur$ and $cur.next$ are empty. If they are not empty, we perform the swap, otherwise we terminate the loop.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the linked list.

/**
 * 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.
# 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.
 * 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
#     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

// 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.
# 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;
    }
}

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

24 - Swap Nodes in Pairs

24. Swap Nodes in Pairs

Description

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24. Swap Nodes in Pairs

Description

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All Problems

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