Leetcode:Swap Nodes in Pairs 单链表相邻两节点逆置

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) {        if(head == NULL || head->next == NULL)            return head;        ListNode dummy(0);        ListNode *tail = &dummy;        ListNode *pre = head;        ListNode *cur = NULL;        ListNode *nxt = NULL;                while(pre != NULL)        {            cur = pre->next;            if(cur != NULL)            {                nxt = cur->next;                cur->next = pre;                pre->next = NULL;                tail->next = cur;                tail = pre;                pre = nxt;            }            else            {                tail->next = pre;                return dummy.next;            }                    }                return dummy.next;    }};