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.
设置个头结点，便于操作，每两个就反转
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *swapPairs(ListNode *head) { 12 if( !head ) return head; 13 ListNode node(-1); //设置头结点 14 node.next = head; 15 ListNode* pre = &node; 16 ListNode* p = head; 17 while( p && p->next ) { //如果当前节点及下一个节点存在 18 pre->next = p->next; 19 p->next = p->next->next; 20 pre->next->next = p; 21 pre = p; 22 p = p->next; 23 } 24 return node.next; 25 } 26 };
原文：http://www.cnblogs.com/bugfly/p/4003149.html