【LeetCode】Swap Nodes in Pairs

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.

code :

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == NULL)
            return NULL;
        if(head->next == NULL)
            return head;

        ListNode *p = head;
        ListNode *q = head->next;
        ListNode *pre = NULL;
        while( q != NULL && q->next != NULL)      // p,q 指向相邻结点一前一后
        {                                        // 注意奇数个结点的情况，判空为第一种
            ListNode *tmp = q->next;
            q->next = p;
            p->next = tmp;
            if( pre == NULL)
            {
                head = q;
                pre = p;
            }
            else
            {
                pre->next = q;
                pre = p;
            }
            p = p->next;
            q = p->next;
        }
        if(pre == NULL)     //只有两个结点
        {
            head = q;
            q->next = p;
            p->next = NULL;
            return head;
        }
        if(q == NULL)       //奇数个结点
        {
            return head;
        }
        q->next = p;        //偶数个结点
        p->next = NULL;
        pre->next = q;
        return head;

    }
};