交换单链表中相邻的两个元素 Swap Nodes in Pairs

题目源自于leetcode。

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. 不允许交换结点的值。空间复杂度必须是O(1)。

思路： 用三个指针进行交换吧。注意开头和结尾的情况。

代码：

/**
 * 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)
            return NULL;
        ListNode *pp, *pre, *p;
        pre = head;
        if(pre->next != NULL)
            head = pre->next;
        else
            return head;
        pp = new ListNode(0);//这个结点没有意义。只是为了给pp赋一个初值。使得第一次交换的情况和后面的代码保持一致
        while(pre != NULL && pre->next != NULL)
        {
            p = pre->next;    
    
            pre->next = p->next;
            p->next = pre;
            pp->next = p;

            pp = pre;
            pre = pre->next;
            
        }
        return head;
    }
};