LeetCode - Swap Nodes in Pairs

Problem :

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.

思路：很简单，就是每次翻转一对，如果不够一对，就保持原来的顺序。奇数个节点的情况要考虑好

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return head;
        ListNode * p = head, * q = head->next, *s = NULL;

        if (q) {
            s = q->next;
            q->next = p;
            p->next = s;
            head = q;
        }
        else
            return head;
        while (s) {
            q = s->next;
            if (q){
                p->next = q;
                p = s;
                s = q->next;
                q->next = p;
                p->next = s;
            }
            else {

                break;
            }


        }
        return head;
    }
};