24. 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.

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這道題不難，但是不能直接交換兩個結點裏面的值，設p,q是要交換的兩個結點，將p指向q->next,相當於刪除q,然後把q插入到p的前面即可。頭結點需要特殊處理一下。

/**
 * 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||head->next==NULL)
      return head;
	ListNode *p,*q,*prior;
	p=head;
	prior=head;
	while (p->next!=NULL)
	{  
		q=p->next;
		p->next=q->next;
		q->next=p;
		if(prior==head)
		{
			head=q;
		}
		else
		{
			prior->next=q;
		}
	    prior=p;
		if(p->next!=NULL)     //如果爲空，回到while判斷的時候會自己退出，
	  	     p=p->next;      //若這裏沒有這個判斷，直接p=p->next，當p->next爲空時，while判斷時再用p->next時會異常。
		 
  }
	return head;
    }
};