README_EN.md

24. Swap Nodes in Pairs

中文文档

Description

Given a linked list, swap every two adjacent nodes and return its head.

 

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

 

Follow up: Can you solve the problem without modifying the values in the list's nodes? (i.e., Only nodes themselves may be changed.)

Solutions

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

/**
 * 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
}

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

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