删除单链表中特定数值范围之外的节点

Question

我正在开发这个程序，它读取文本文件并从文本文件中抓取数据并将其插入到链表的节点中。

除了删除节点外，我的整个程序都运行良好。我正在从文本文件中过滤数据，所以我只需要打印出具有特定范围内的值的数据。我可以用 if() 语句来做到这一点，它工作得很好，但这不是我想要的结果。

我想删除指定范围之外的节点并释放它们正在使用的内存。我写了几行代码试图做到这一点，但它最终会删除整个列表。因此，如果有人能指出我正确的方向并告诉我我做错了什么，那就太好了！

#include <fstream>
#include <iostream>
using namespace std;

struct Employee
{
    string firstN;
    string lastN;
    float salary;
    float bonus;
    float deduction;

    Employee *link;
};

typedef Employee* EmployPtr;
void insertAtHead( EmployPtr&, string, string, float, float,float );
void insert( EmployPtr&, string, string, float, float,float );
float netSalary( EmployPtr& );

int main()
{
//Open file
fstream in( "payroll.txt", ios::in );

//Read lines
string first, last;
float salary, bonus, deduction;
EmployPtr head = new Employee;

//Inserts all the data into a new node in the linked list, creating a new node each time the loop executes.
while( in >> first >> last >> salary >> bonus >> deduction)
    insertAtHead (head, first, last, salary, bonus, deduction);

//Close file
in.close();

cout << "-Salary in the range of ($45,000 - $60,000)-\n" << "Printed in format: First Name, Last Name, Salary, Bonus, Deduction, Net Salary.\n";

EmployPtr iter, temp;
for(iter = head; iter!= NULL; iter = iter->link)
{
    temp = head;
    //Deletes nodes outside of range.
    while(netSalary(iter)<45000 || netSalary(iter)>60000)
    {
        EmployPtr nodeToDelete = temp;
        temp = temp->link;
        delete nodeToDelete;
    }

    cout << iter->firstN << ", " << iter->lastN << ", " << iter->salary << ", " << iter->bonus << ", " << iter->deduction << ", " << netSalary(iter) <<endl;

}
    return 0;
}

    //Based off of the input values, this function will create a new node and insert it at the beginning of the linked list. This function ONLY allows insertion at the beginning of the list and no where else.
 void insertAtHead(EmployPtr& head, string firstValue, string lastValue,
            float salaryValue, float bonusValue,float deductionValue)
{
    EmployPtr tempPtr= new Employee;

    tempPtr->firstN = firstValue;
    tempPtr->lastN = lastValue;
    tempPtr->salary = salaryValue;
    tempPtr->bonus = bonusValue;
    tempPtr->deduction = deductionValue;

    tempPtr->link = head;
    head = tempPtr;
}

//Based off of the input values, this function creates a new node and inserts it AFTER the node provided in the argument.
void insert(EmployPtr& afterNode, string firstValue, string lastValue,
        float salaryValue, float bonusValue,float deductionValue)
{
    EmployPtr tempPtr= new Employee;


    tempPtr->firstN = firstValue;
    tempPtr->lastN = lastValue;
    tempPtr->salary = salaryValue;
    tempPtr->bonus = bonusValue;
    tempPtr->deduction = deductionValue;

    tempPtr->link = afterNode->link;
    afterNode->link = tempPtr;
}

//This function calculates a net salary based off of the salary, bonus, and deduction variables of the input node.
float netSalary(EmployPtr& node)
{
    float netSalary, newDeduction;

    newDeduction = ((node->salary) + (node->bonus)) * (node->deduction);
    netSalary = (node->salary + node->bonus) - newDeduction;

    return netSalary;
}

编辑：将 && 改回 ||还是有问题。

编辑 #2：解决方案

while(netSalary(iter)<45000 || netSalary(iter)>60000)
        {
            EmployPtr nodeToDelete = new Employee;
            nodeToDelete = iter;
            iter = iter->link;
            delete nodeToDelete;
        }

Answer 1

这一行就在这里：

while(netSalary(iter)<45000 && netSalary(iter)>60000)

我相信您的条件应该是 OR (||)。一个值同时小于 45000 和大于 60000 是没有意义的。

给定值 25000，它将小于 45000，但不大于 60000，因此不会删除任何内容。

编辑： 或许可以尝试以下方式：

for (iter = head; iter != NULL; iter = iter->link) 
{
    cout << iter->salary; // so you can see what node it's looking at
    if (netSalary(iter) < 45000 || netSalary(iter) > 60000)
    {
        EmployPtr nodeToDelete = iter;
        iter = iter->link; // difference here is that you're explicitly moving the iter forward
        delete nodeToDelete;
    } 
}

Answer 2

我想你想要的条件是

while(netSalary(iter) >= 45000  &&  netSalary(iter) <= 60000)

我说想想是因为我没有看到你真正想要过滤掉的东西的声明。

Answer 3

您应该更改 while 条件，作为开始。 '&&' 应该是 '||'因为 '

编辑：

好的，我相信问题在于使用“iter”的 while 循环。一旦迭代器在 while 循环中与您的条件匹配，您就不会在此之后将迭代器向前移动（因为您不会返回到 for 循环），因此会删除 while 循环内的列表的其余部分。尝试将时间更改为“如果”，看看你会得到什么。

Answer 4

最新的解决方案应该解决主要问题（删除整个列表，因为循环以 head 而不是 iter 开始），但您可能仍会遇到另一个问题。如果最后一个元素被删除，下次检查循环条件时，netSalary 将在一个空指针上调用（因为一旦它前进到 iter->link，iter 就会为空）。此外，尝试修改该循环以考虑空指针可能会导致外部 for 循环尝试访问空指针的链接成员。

我可以建议的最简单的解决方案是修改代码以仅使用一个 while 循环和条件，如下面的代码所示：

EmployPtr iter = head, temp;
while(iter!= NULL)
{
    if(netSalary(iter)<45000 || netSalary(iter)>60000)
    {
        // bad node, delete and advance
        EmployPtr nodeToDelete = iter; 
        iter = iter->link;
        delete nodeToDelete;
    } 
    else 
    {
        // good node, write and advance
        cout << iter->firstN << ", " << iter->lastN << ", " << iter->salary << ", " << iter->bonus << ", " << iter->deduction << ", " << netSalary(iter) <<endl;
        iter = iter->link;
    }
}