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Given a linked list, swap every two adjacent nodes and return its head.
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.
LeetCode上的原题，请参见我之前的博客Swap Nodes in Pairs。
解法一：
class Solution { public: /** * @param head a ListNode * @return a ListNode */ ListNode* swapPairs(ListNode* head) { ListNode *dummy = new ListNode(-1), *pre = dummy; dummy->next = head; while (pre->next && pre->next->next) { ListNode *t = pre->next; pre->next = t->next; t->next = t->next->next; pre->next->next = t; pre = t; } return dummy->next; } };
解法二：
class Solution { public: /** * @param head a ListNode * @return a ListNode */ ListNode* swapPairs(ListNode* head) { if (!head || !head->next) return head; ListNode *t = head->next; head->next = swapPairs(head->next->next); t->next = head; return t; } };
[LintCode] Swap Nodes in Pairs 成对交换节点
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原文地址：http://www.cnblogs.com/grandyang/p/5648424.html
