【Swap Nodes in Pairs】cpp

最新推荐文章于 2025-09-11 19:30:36 发布

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最新推荐文章于 2025-09-11 19:30:36 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/xbf9xbf/p/4468079.html

本文提供了一种链表中相邻节点的交换算法实现，通过使用常数空间复杂度完成节点交换，而非修改链表中的值。该算法适用于任何单向链表，并提供了完整的C++代码实现及解析。

题目：

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

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
            if (!head || !(head->next) ) return head;
            ListNode dummy(-1);
            dummy.next = head;
            ListNode *prev = &dummy;
            ListNode *curr = head;
            while ( curr && curr->next )
            {
                prev->next = curr->next;
                curr->next = curr->next->next;
                prev->next->next = curr;
                prev = curr;
                curr = curr->next;
            }
            return dummy.next;
    }
};

Tips:

链表基本操作，动手画图，直接出来。

===================================

第二次过这道题，一次AC了。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
            ListNode dummpy(-1);
            dummpy.next = head;
            ListNode* pre = &dummpy;
            ListNode* curr = head;
            while ( curr && curr->next )
            {
                pre->next = curr->next;
                curr->next = curr->next->next;
                pre->next->next = curr;
                pre = curr;
                curr = curr->next;
            }
            return dummpy.next;
    }
};

转载于:https://www.cnblogs.com/xbf9xbf/p/4468079.html