c++ - 遍历插入 STL 向量值的链表

Question

我正在尝试迭代一个向量并将数据放入链表节点......我知道我可以将 STL 迭代器用于向量，但是我可以用什么来循环链表？我不认为我可以使用 STL 列表迭代器，对吧？

列表.h

class List {
public:
    List();
    void addNode(int addData);
    void deleteNode(int delData);
    void printList();

private:
    typedef struct Node {
        int data;
        Node* next;
    }* nodePtr;

    nodePtr head;
    nodePtr curr;
    nodePtr temp;
};

列表.cpp

List::List() {
    head = NULL;
    curr = NULL;
    temp = NULL;
}

void List::addNode(int addData){
    nodePtr n = new Node;
    n->next = NULL;
    n->data = addData;

    if(head != NULL) {
        curr = head;
        while (curr->next != NULL) {
            curr = curr->next;
        }
        curr->next = n;
    }
    else {
        head = n;
    }
}

void List::deleteNode(int delData) {
    nodePtr delPtr = NULL;
    temp = head;
    curr = head;
    while(curr != NULL && curr->data != delData) {
        temp = curr;
        curr = curr->next;
    }
    if(curr == NULL) {
        cout << delData << " was not in the list.\n";
        delete delPtr;
    }
    else {
        delPtr = curr;
        curr = curr->next;
        temp->next = curr;
        if(delPtr == head) {
            head = head->next;
            temp = NULL;
        }
        delete delPtr;
    }
}

void List::printList() {
    curr = head;
    while(curr !=NULL) {
        cout << curr->data << endl;
        curr= curr->next;
    }
}

主文件

#include <cstdlib>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <iterator>

int main(int argc, char** argv) {
    cout << "Enter some integers, space delimited:\n";
    string someString;
    getline(cin, someString);

    istringstream stringStream( someString );
    vector<int> integers;
    int n;
    while (stringStream >> n)
        List listOfInts;
    listOfInts.addNode(/* stuff in here*/)
    integers.push_back(n);

    return 0;
}

Answer 1

您不需要遍历链表。用于addNode将项目添加到链接列表。

vector<int> vec;
...
List list;
for (vector<int>::iterator i = vec.begin(); i != vec.end() ++i)
    list.addNode(*i);

这里的所有都是它的。