LeetCode-24. Swap Nodes in PairsLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/swap-nodes-in-pairs/

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

 【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) {}
 * };
 */

1. 循环

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode *dummy = new ListNode(-1), *pre = dummy;
        dummy->next = head;
        while (pre->next != nullptr && pre->next->next != nullptr) {
            ListNode *l2 = pre->next->next;
            pre->next->next = l2->next;
            l2->next = pre->next;
            pre->next = l2;
            pre = pre->next->next;
        }
        return dummy->next;
    }
};

2. 递归

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if (head == nullptr || head->next == nullptr) {return head;}
        ListNode* newHead = head->next;
        head->next = swapPairs(head->next->next);
        newHead->next = head;
        return newHead;
    }
};

【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; }
 * }
 */

1. 循环

class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode dummy = new ListNode(-1), pre = dummy;
        dummy.next = head;
        while (pre.next != null && pre.next.next != null) {
            ListNode l2 = pre.next.next;
            pre.next.next = l2.next;
            l2.next = pre.next;
            pre.next = l2;
            pre = pre.next.next;
            
        }
        return dummy.next;
    }
}

2.递归

class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {return head;}
        ListNode newHead = head.next;
        head.next = swapPairs(head.next.next);
        newHead.next = head;
        return newHead;
    }
}