24. Swap Nodes in Pairs leetcode(list)

1. 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.
   方法一、递归的解法

ListNode* swapPairs(ListNode* head) {        if(head == NULL || head->next == NULL)            return head;        ListNode* newHead = head->next;        head->next = swapPairs(head->next->next);        newHead->next = head;        return newHead;    }

方法二、非递归的解法

ListNode* swapPairs(ListNode* head) {        if(NULL == head || NULL == head->next)        {            return head;        }        ListNode* first = head;        ListNode* second = first->next;        ListNode* pre;        int count = 0;        while(first && second)        {            first->next = second->next;            second->next = first;            if(count == 0) //如果是最开始的两个节点，则需要更改头节点            {                head = second;            }            else //如果不是开始的两个节点，则在交换第二个节点时，需要将第二个节点插在上一轮交换的最后一个节点之后，用pre保存            {                pre->next = second;            }            pre = first; //更新pre节点            first = first->next;            if(first) //注意如果first已经为NULL了，如果不加入if判断则会出现runtime 的错误            {                second = first->next;            }            count++;        }        return head;    }