［LeetCode］24. Swap Nodes in Pairs

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.

主要是要保存头结点以及交换的第二个结点后一个结点。

/**
 * 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 || !head->next) return head;
        ListNode first(0);
        ListNode *pre = &first;
        ListNode *cur1 = head, *cur2 = head->next;
        while(cur1 && cur2){
            ListNode* node2Next = cur2->next;
            pre->next = cur2;
            cur2->next = cur1;
            cur1->next = node2Next;
            pre = cur1;
            if (!node2Next)break;
            cur1 = node2Next;
            cur2 = node2Next->next;
        }
        return (&first)->next;
    }
};