[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.

思路：每次交换cur1和cur2，使用dummy保证了pre的初始赋值的正确性，同时，注意next有可能为空。

/**
 * 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 dummy(-1);
            dummy.next = head;

            ListNode* pre = &dummy;
            ListNode* cur1 = head;
            ListNode* cur2 = head->next;
            ListNode* next = NULL;

            // swap cur1 and cur2
            while(cur1 && cur2)
            {   
                next = cur2->next;
                pre->next = cur2;
                cur2->next = cur1;
                cur1->next = next;

                pre = cur1;
                cur1 = next;
                // next may be NULL
                if(next == NULL)
                    break;
                else
                    cur2 = next->next;
            }   
                
            return dummy.next;
        }   
};