LeetCode 24. Swap Nodes in Pairs ***** 双节点交换

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.

题意

给定一个链表，交换相邻的两个结点，返回交换之后的头结点

注意

是交换相邻的元素，也是说，两两交换，交换之后不在参与，当不满足两个元素不用交换

思路

设置四个指针，pre，node1，node2，next。关键在于pre指针。

代码

class Solution {public:    ListNode* swapPairs(ListNode* head) {        ListNode *virtualNode = new ListNode(0);        virtualNode->next = head;        ListNode *pre = virtualNode;        while(pre->next!=NULL&&pre->next->next != NULL)        {            ListNode* node1 = pre->next;            ListNode* node2 = node1->next;            ListNode* next = node2->next;            node2->next = node1;            node1->next = next;            pre->next = node2;            pre = node1;        }        ListNode* resNode = virtualNode->next;        delete virtualNode;        return resNode;    }};

结果