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Description:
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.
Code:
1 ListNode* swapPairs(ListNode* head) { 2 if (!head) 3 return NULL; 4 ListNode *preNode = new ListNode(0); 5 preNode ->next = head; 6 7 ListNode *m_preNode = preNode; 8 ListNode *p = head; 9 ListNode *q = p->next; 10 11 while (p&&q) 12 { 13 preNode->next = q; 14 p->next = q->next; 15 q->next = p; 16 17 preNode = p; 18 p = preNode->next; 19 //要注意检测p是否为空值，否则对q赋值就会出现错误 20 if (p) 21 q = p->next; 22 else 23 q = NULL; 24 } 25 return m_preNode->next; 26 }
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原文地址：http://www.cnblogs.com/happygirl-zjj/p/4575664.html
