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.
在不新建一个链表的情况下，交换链表相邻节点。
交换相邻节点A->B，A是head.
ListNode* p = head; head = head->next; p->next = p->next->next; head->next = p;
整体代码：
1 class Solution { 2 public: 3 ListNode* swapPairs(ListNode* head) { 4 if (!head || !head->next) 5 return head; 6 ListNode* p = head; 7 head = head->next; 8 p->next = p->next->next; 9 head->next = p; 10 ListNode* q = p->next; 11 while (q && q->next) { 12 ListNode *temp = q; 13 q = q->next; 14 temp->next = temp->next->next; 15 q->next = temp; 16 p->next = q; 17 p = q->next; 18 q = p->next; 19 } 20 return head; 21 } 22 };
