LeetCode-24. Swap Nodes in Pairs [C++][Java]

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;
    }
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

贫道绝缘子

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值