栈（链表实现）

1.思路
节点结构体为：一个int数值，一个指向下一个节点的指针Struct* next。设定一个链表头节点p，每次入栈的节点都放在p点的下一个节点，前面的节点则依次往后推移一位；每次出栈时将p->next对应节点值输出，并将p->next指向p->next->next；当p->next为nullptr则表明栈为空；当p为nullptr，表明栈不存在。
2.代码

#include <iostream>using namespace std;struct Stack {    int value;    struct Stack *next;};Stack* CreateStack() {    Stack* p = (Stack*)malloc(sizeof(Stack));    p->value = 0;    p->next = nullptr;    return p;}bool isEmpty(Stack* p) {    if(p == nullptr) {        cout << "stack creats failed" << endl;        return true;    }    return p->next == nullptr;}Stack* PushStack(Stack* p, int value) {    if(p == nullptr) {        cout << "stack creats failed" << endl;        return nullptr;    }    Stack* tmp = (Stack*)malloc(sizeof(Stack));    tmp->value = value;    tmp->next = p->next;    p->next = tmp;    return p;}int PopStack(Stack* p) {    if(p == nullptr) {        cout << "stack creats failed" << endl;        return -100000000;    }    Stack* tmp = p->next;    p->next = tmp->next;    int res = tmp->value;    free(tmp);    return res;}void Destroy(Stack* p) {    while(!isEmpty(p)) {        PopStack(p);    }    free(p);    p = nullptr;}int main() {    Stack* p = CreateStack();    cout << "push 0-9" << endl;    for(int i = 0; i < 10; ++i) {        p = PushStack(p, i);    }    cout << "pop stack until it's empty" << endl;    while(!isEmpty(p)) {        cout << PopStack(p) << endl;    }    cout << "push 22" << endl;    p = PushStack(p, 22);    cout << "push 6" << endl;    p = PushStack(p, 6);    cout << "push 11" << endl;    p = PushStack(p, 11);    cout << "pop" << endl;    cout << PopStack(p) << endl;    cout << "pop" << endl;    cout << PopStack(p) << endl;    cout << "push 8" << endl;    p = PushStack(p, 8);    cout << "pop" << endl;    cout << PopStack(p) << endl;    cout << "pop" << endl;    cout << PopStack(p) << endl;    Destroy(p);    return 0;}

3.运行结果