C++ 删除链接列表中的节点函数

我有一个程序，可以询问用户希望删除哪个项目。然后，delete函数应成功删除该节点。我知道我需要从上一个节点获取地址，并使其指向我要删除的节点之后的节点，然后删除悬挂节点。我在理解语法方面遇到了一些困难。这是我的密码。我省略了不必要的功能

#include <iostream> 
#include <cstddef> 
#include <string> 
using namespace std;

struct Node 
{
    string item; 
    int count; 
    Node *link; 
};

typedef Node* NodePtr;

void insert(NodePtr after_me, string an_item, int a_number); 
void list_remove(NodePtr& head, string an_item);
void head_insert(NodePtr& head, string an_item, int a_number); 
void show_list(NodePtr& head); 
NodePtr search(NodePtr head, string target);

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因此，由于您没有真正展示您将要尝试的内容，我无法真正说出哪个“语法”让您感到特别困惑。只是一个猜测

如果我理解正确，这归结为您想知道如何找到放置在您要移除的项目之前的项目，以便在必要时对其进行操作，对吗？
然后自己迭代列表，而不是使用

search（）

方法：

void list_remove(NodePtr& head, string remove_item)
{
    NodePtr remove_ptr; //pointer to the node that is being removed

    remove_ptr = search(head, remove_item);
    if(remove_ptr != NULL) // indicates the item is part of the list and 
                           // we can remove it    
    {
         NodePtr predecessorNode = NULL;
         // Iterate the list to find the predecessor of remove_ptr
         for(NodePtr curNode = head; curNode != NULL; curNode = curNode->link)
         {
             if(curNode->link == remove_ptr) // Indicates that curNode is the
                                             // predecessor of remove_ptr
             {
                 predecessorNode = curNode;  // store it and ...
                 break;                      // ... exit the loop
             }
         }

         if(predecessorNode != NULL) // We have found a predecessor
         {
             // Now you have the predecessor for remove_ptr, 
             // manipulate it as necessary
         }
         else // indicates the item to remove is the head of the list
         {
             // You might need to manipulate head, depending on how your 
             // linked list is organized (set to remove_ptr->link most likely)
         }
    }
}

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要删除，还需要先删除节点，然后再删除节点。 这些步骤是：

// find the preceding_ptr
NodePtr preceding_ptr = head;
while(preceding_ptr != NULL && preceding_ptr->link != remove_ptr){
    preceding_ptr = preceding_ptr->link;
}

// now preceding_ptr is set, make it skip remove_ptr
preceding_ptr->link = remove_ptr->link;

// free up memory from remove_ptr

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不清楚您在问什么。我不确定应该向删除功能添加什么才能使其正常工作。语法令人困惑。请澄清哪些“语法”特别让您困惑，而不是将所有代码都扔给我们！顺便说一句：你有没有办法让它成为一个双链接列表，这样可以更容易地插入和删除？另外一点：改进代码格式！我只编辑了一些基础知识…仅供参考，这是您的列表删除方法的主体。

// Uses cstddef: 
void head_insert(NodePtr& head, string an_item, int a_number) 
{ 
    NodePtr temp_ptr; 
    temp_ptr = new Node;

    temp_ptr -> item = an_item; 
    temp_ptr -> count = a_number;

    temp_ptr->link = head; 
    head = temp_ptr; 
}

//Uses iostream and cstddef: 
void show_list(NodePtr& head) 
{ 
    NodePtr here = head;

    while (here != NULL) 
    { 
        cout << here-> item << "\t"; 
        cout << here-> count << endl; 
        here = here->link; 
    } 
}

NodePtr search(NodePtr head, string target) 
{ 
    // Point to the head node
    NodePtr here = head;

    // If the list is empty nothing to search 
    if (here == NULL) 
    { 
         return NULL; 
    }
    // Search for the item 
    else 
    { 
         // while you have still items and you haven't found the target yet 
         while (here-> item != target && here->link != NULL) 
             here = here->link;

         // Found the target, return the pointer at that location 
         if (here-> item == target) 
             return here;
         // Search unsuccessful, return Null 
         else 
             return NULL; 
    } 
}

void list_remove(NodePtr& head, string remove_item)
{
    NodePtr remove_ptr; //pointer to the node that is being removed

    remove_ptr = search(head, remove_item);
    if(remove_ptr != NULL) // indicates the item is part of the list and 
                           // we can remove it    
    {
         NodePtr predecessorNode = NULL;
         // Iterate the list to find the predecessor of remove_ptr
         for(NodePtr curNode = head; curNode != NULL; curNode = curNode->link)
         {
             if(curNode->link == remove_ptr) // Indicates that curNode is the
                                             // predecessor of remove_ptr
             {
                 predecessorNode = curNode;  // store it and ...
                 break;                      // ... exit the loop
             }
         }

         if(predecessorNode != NULL) // We have found a predecessor
         {
             // Now you have the predecessor for remove_ptr, 
             // manipulate it as necessary
         }
         else // indicates the item to remove is the head of the list
         {
             // You might need to manipulate head, depending on how your 
             // linked list is organized (set to remove_ptr->link most likely)
         }
    }
}

// find the preceding_ptr
NodePtr preceding_ptr = head;
while(preceding_ptr != NULL && preceding_ptr->link != remove_ptr){
    preceding_ptr = preceding_ptr->link;
}

// now preceding_ptr is set, make it skip remove_ptr
preceding_ptr->link = remove_ptr->link;

// free up memory from remove_ptr