[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>