24. Swap Nodes in Pairs

最新推荐文章于 2025-09-08 10:56:58 发布

转载最新推荐文章于 2025-09-08 10:56:58 发布 · 52 阅读

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原文链接：https://my.oschina.net/u/1240964/blog/825486

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本文介绍了一种链表中相邻节点的交换算法实现，通过使用常数空间而非修改列表值的方式，达到节点位置交换的目的。示例展示了如何将1->2->3->4的链表转换为2->1->4->3的形式。

2019独角兽企业重金招聘Python工程师标准>>>

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.

/**
 * 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) {
        struct ListNode *tmphead,*ptr1,*ptr2;
        tmphead = new ListNode(0);
        ptr1 = new ListNode(0);
        ptr2 = new ListNode(0);
        tmphead->next = head;
        head = tmphead;

        while (head->next != NULL && head->next->next != NULL){
            ptr1 = head->next,ptr2 = head->next->next;
            //交换位置
            head->next = ptr2;     
            ptr1->next = ptr2->next;
            ptr2->next = ptr1;
            //更新待交换位置，由于ptr1更换到后位置，所以head=ptr1
            head = ptr1;
   
        }
        return tmphead->next;
    }
};

转载于:https://my.oschina.net/u/1240964/blog/825486