使用c++中的单链表完成问题（1）从键盘输入10个整数，产生顺序表，并输入结点值

（1）从键盘输入10个整数，产生顺序表，并输入结点值
（2）从键3盘输入1个整数，在顺序表中查找该结点的位置，若找到，输出结点的位置，若找不到，则显示"找不到"
→用单链表实现←

首先顺序表不是链表哦，顺序表是内存连续的一种线性表，比如数组。第一个到底是产生顺序表还是链表呢？

    //实现链表倒置,如链表存储的是1,2,3,倒置后为3,2,1 #include <iostream> using namespace std; //定义链表 typedef struct node { int data; node * next; }node; //创建链表 node* createlist(int &n) { node *tmp, *head = null, *tail = null; int num; cin>>num; head = new node; if(head == null) {   cout<<"no memory available!";   return null; } head->data = num; head->next = null; tail = head; for(int i=0; i<n-1; i++) {   cin>>num;   tmp = new node;   if(tmp == null)   {    cout<<"no memory available!";    return null;   }   tmp->data = num;   tmp->next = null;   tail->next = tmp;   tail = tmp; } return head; } //链表倒置 node* inversionlist(node *head) { if(head == null || head->next == null) {   return head; } node *connode = head->next; node *curnode = connode->next; head->next = null; while(curnode != null) {   connode->next = head;   head = connode;   connode = curnode;   curnode = curnode->next; } connode->next = head; head = connode; return head; }

//打印     void printlist(node *head) { node *curnode = head; while(curnode != null) {   cout<<curnode->data<<"\t";   curnode = curnode->next; } } //删除链表,释放内存 void deleteallnode(node *head) { node *p = null; while(head != null) {   p = head;   head = head->next;   delete p;   p = null; } } void main() { int n; cout<<"please input the number of node:"; cin>>n; node *head = createlist(n); node *newhead = inversionlist(head); printlist(newhead); deleteallnode(newhead);

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