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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.
题意：交换一个链表中相邻的两个元素，不能只交换结点元素值，要进行结点的操作。
思路：使用3个指针，分别记录当前指针，当前指针的前驱，当前指针的后继，然后进行交换操作。
代码：
public ListNode swapPairs(ListNode head) { if(head==null) return head; ListNode pPrePre = null; ListNode pPre = null; ListNode p = head; while(p!=null&&p.next!=null){ pPre = p; p = p.next; ListNode pNext = p.next; if(pPre==head) head=p; if(pPrePre!=null) pPrePre.next = p; p.next = pPre; pPre.next = pNext; pPrePre = pPre; p=pNext; } return head; }
LeetCode(24): Swap Nodes in Pairs
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原文地址：http://www.cnblogs.com/Lewisr/p/5134482.html
