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.
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 == NULL || head->next == NULL) return head; 13 14 ListNode **prev = &head; 15 while ((*prev) != NULL && ((*prev)->next) != NULL) { 16 ListNode *pcurr = *prev; 17 ListNode *pnext = (*prev)->next; 18 pcurr->next = pnext->next; 19 pnext->next = pcurr; 20 *prev = pnext; 21 prev = &(pcurr->next); 22 } 23 24 return head; 25 } 26 };
[LeetCode] Swap nodes in pairs
原文：http://www.cnblogs.com/vincently/p/4060154.html