[leetcode 24]Swap Nodes in Pairs-----成对翻转链表中的节点

Question:

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.

分析：

比较简单，别忘了记录前面的节点即可，避免丢节点。

代码如下：

<span style="font-size:14px;">/** * 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* l = head;        ListNode* r = head->next;        ListNode* p;        head = r;        l->next = r->next;        r->next = l;        while(l->next != NULL && l->next->next != NULL){            p = l;            l = l->next;            r = l->next;            l->next = r->next;            r->next = l;            p->next = r;        }                return head;    }};</span>