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题目：
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.
交换链表相邻结点
思路：只要保存头结点，然后理清交换结点位置就好了
例如链表现在是1->2->3->4,head指向1，新建一个结点root，root->next指向head
第一次循环：pre指向root，要交换12，首先保存2的下一个结点3，接下来让pre->next为2,2的next为1,1的next为刚才保存的3，然后让head指向3，pre指向1就好了，然后继续
代码：
/** * 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 *root = new ListNode(0); root->next = head; ListNode *pre = root; while(head && head->next) { ListNode* tmp = head->next->next; pre->next = head->next; pre->next->next = head; head->next = tmp; pre = head; head = head->next; } return root->next; } };
[LeetCode24]Swap Nodes in Pairs
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原文地址：http://www.cnblogs.com/zhangbaochong/p/5152169.html
