Question

我得到了这两个结构

struct CamelZombie{
    int hp;
    int attack;
    CamelZombie *next;
};

struct list_of_cz{
    CamelZombie *head;
};

我已经创建了一个函数来创建具有给定值的链表：

void createCamelZombie(list_of_cz *&pZ, int z_hp, int z_attack, int N){
    pZ = new list_of_cz;
    pZ->head->hp = z_hp;
    pZ->head->attack = z_attack;
    CamelZombie *temp1 = pZ->head;
    CamelZombie *temp2 = NULL;
    for (int i = 0; i < N - 1  ; i++){
        temp2 = new CamelZombie;
        temp2->hp = z_hp;
        temp2->attack = z_attack;
        temp1->next = temp2;
        temp1 = temp2;
    }
}

然后我把它放在这样的功能主体中，但是然后推进了，不知道为什么。

list_of_cz *pZ = NULL;
createCamelZombie(pZ, z_hp, z_attack, N);
    while (pList->head != NULL && pZ != NULL){
        atPlant(numPlant(pList) - 1, pList)->hp -= pZ->head->attack;
        if (atPlant(numPlant(pList) - 1, pList)->hp <= 0) deletePlant(numPlant(pList) - 1, pList);
        int count = 0;
        CamelZombie *z_temp;
        z_temp = pZ->head;
        while (z_temp){
            if (count == 0) z_temp->hp -= allPlantAttack(pList, numPlant(pList) - 1);
            else z_temp->hp -= allLaserAttack(pList); //trouble right here
            if (z_temp->hp <= 0) deleteCamelZombie(pZ, count);
            z_temp = z_temp->next;
            count++;
        }

似乎我在写void createCamelZombie()时错过了一些东西，因为编译器告诉我z_temp->hp没有值。请帮帮我！

Answer 1

最好使用现有容器，例如std::vector或std::list

#include <iostream>
#include <string>
#include <list>

struct CamelZombie{
  std::string name; //added for demonstration purposes
  int hp;
  int attack;
  //pointer to next zombie not required
};

std::list<CamelZombie> createCamelZombie2(int z_hp, int z_attack, int N) {

  std::list<CamelZombie> result;

  for (int i = 0; i < N; i++){

    CamelZombie newZombie;
    newZombie.name = "Zombie"+std::to_string(i);
    newZombie.hp = z_hp;
    newZombie.attack = z_attack;
    newZombie.next = NULL;

    result.push_back(newZombie);
  }

  return result;
}

使用这样的代码。

int main() {
  std::list<CamelZombie> listOfZombies2 = createCamelZombie2(10,20,10);

  for(std::list<CamelZombie>::iterator list_iter = listOfZombies2.begin(); 
      list_iter != listOfZombies2.end(); list_iter++)
  {
    std::cout<<list_iter->name<<std::endl;
  }

}

如果您真的想使用自己的链接列表，请尝试下面的代码。

不需要列表的单独结构（list_of_cz）。每个僵尸链接到下一个僵尸。所以只需保持指向第一个僵尸的指针。
createCamelZombie函数返回指向列表中第一个僵尸的指针（不需要使用函数参数（list_of_cz *＆amp; pZ）来获取僵尸列表）
太多的下划线和Z使代码难以阅读。
如果你使用指针，你需要自己清理内存。

struct CamelZombie{
  std::string name; //added for demonstration purposes
  int hp;
  int attack;
  CamelZombie *next;
};

CamelZombie* createCamelZombie(int z_hp, int z_attack, int N){

  CamelZombie *result = NULL;
  CamelZombie *work = NULL; //keep track of the last node in the list

  for (int i = 0; i < N; i++){

    //create new zombie
    CamelZombie *newZombie = new CamelZombie();
    newZombie->name = "Zombie"+std::to_string(i);
    newZombie->hp = z_hp;
    newZombie->attack = z_attack;
    newZombie->next = NULL;

    if (result==NULL) {
      result = newZombie;
      work =result;
    } else {
      work->next = newZombie;
      work = newZombie;
    }
  }

  return result;
}

如何使用代码的示例。

int main() {

  CamelZombie *listOfZombies = createCamelZombie(10,20,10);

  CamelZombie *work = listOfZombies;

  // print zombie names to screen ---------
  while (work!=NULL) {
    std::cout << work->name << std::endl;
    work = work->next;
  }

免费记忆。

  work = listOfZombies;
  while (work!=NULL) {
    CamelZombie *temp =work->next;
    delete work;
    work = temp;
  }

如何在C ++中创建链表？

1 个答案: