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.
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ListNode* swapPairs(ListNode* head) { if (head == nullptr || head->next == nullptr) return head; ListNode *pre = head; ListNode *cur = pre->next; ListNode *net = cur->next; head = cur; while (true) { cur->next = pre; if (net == nullptr || net->next == nullptr) { pre->next = net; break; } pre->next = net->next; pre = net; cur = pre->next; net = cur->next; } return head; }
惊人的解法，用双重指针，开阔思路，本以为自己可以随心所欲使用双指针，看来还是差得远
ListNode *swapPairs(ListNode *head) { ListNode **p = &head; while (*p && (*p)->next) { ListNode *t = (*p)->next; (*p)->next = t->next; t->next = *p; *p = t; p = &(*p)->next->next; } return head; }
原文：http://www.cnblogs.com/sdlwlxf/p/5122770.html