面试经典题单链表反转

struct node{node* next;T value;
};

node* reverse(node*& head)
{if ( (head &＃61;&＃61; Null) || (head->next &＃61;&＃61; Null) ) return head;// 边界检测node* pNext &＃61; Null;node* pPrev &＃61; head;// 保存链表头节点node* pCur &＃61; head->next;// 获取当前节点while (pCur !&＃61; Null){pNext &＃61; pCur->next;// 将下一个节点保存下来pCur->next &＃61; pPrev;// 将当前节点的下一节点置为前节点pPrev &＃61; pCur;// 将当前节点保存为前一节点pCur &＃61; pNext;// 将当前节点置为下一节点}head->next &＃61; NULL;head &＃61; pPrev; return head;
}

node* reverse( node* pNode, node*& head){if ( (pNode &＃61;&＃61; Null) || (pNode->next &＃61;&＃61; Null) ) // 递归跳出条件{head &＃61; pNode; // 将链表切断&＃xff0c;否则会形成回环return pNode;}node* temp &＃61; reserve(pNode->next, head);// 递归temp->next &＃61; pNode;// 将下一节点置为当前节点&＃xff0c;既前置节点return pNode;// 返回当前节点}