数据结构-链表习题（C++）

程序设计题: 单链表实验
1.从键盘输入若干大于0的整数，用这些整数构造一个单链表.当用户输入小于等于0的值时创建链表结束并在终端打印输出这个链表。
2.在链表中查找某结点，如果能找到输出这个结点是第几个结点，如果找不到，输出:找不到此结点。
3.删除指定结点: 如果指定的被删除结点存在就删除它,然后打印:已经删除，如果不存在，输出信息: 链表中没有此结点，无法删除。
4.删除重复结点:如果链表中有重复的元素，删除重复的元素，使得所有值最多只出现一次，例如，如果链表中有3个结点的值是100，那么需要删除两个值为100的结点。只保留一个值为100的结点。删除重复结点后打印输出整个链表

1------------------------------------------------------
#include <iostream>
using namespace std;

struct Node {
    int data;
    Node* next;
};

Node* createLinkedList() {
    Node* head = nullptr;
    Node* tail = nullptr;
    int input;
    while (true) {
        cin >> input;
        if (input <= 0) {
            break;
        }
        Node* newNode = new Node;
        newNode->data = input;
        newNode->next = nullptr;
        if (head == nullptr) {
            head = newNode;
            tail = newNode;
        } else {
            tail->next = newNode;
            tail = newNode;
        }
    }
    return head;
}

void printLinkedList(Node* head) {
    Node* current = head;
    while (current != nullptr) {
        cout << current->data << " ";
        current = current->next;
    }
}

int main() {
    Node* head = createLinkedList();
    printLinkedList(head);
    return 0;
}

2------------------------------------------------------
int findNodePosition(Node* head, int value) {
    Node* current = head;
    int position = 1;
    while (current != nullptr) {
        if (current->data == value) {
            return position;
        }
        current = current->next;
        position++;
    }
    return -1; // Node not found
}
int main() {
    Node* head = createLinkedList();
    int valueToFind;
    cin >> valueToFind;
    int position = findNodePosition(head, valueToFind);
    if (position != -1) {
        cout << "The node is at position: " << position << endl;
    } else {
        cout << "Node not found" << endl;
    }
    return 0;
}
3------------------------------------------------------
int findNodePosition(Node* head, int value) {
    Node* current = head;
    int position = 1;
    while (current != nullptr) {
        if (current->data == value) {
            return position;
        }
        current = current->next;
        position++;
    }
    return -1; // Node not found
}

int main() {
    Node* head = createLinkedList();
    int valueToFind;
    cin >> valueToFind;
    int position = findNodePosition(head, valueToFind);
    if (position != -1) {
        cout << "The node is at position: " << position << endl;
    } else {
        cout << "Node not found" << endl;
    }
    return 0;
}
4------------------------------------------------------
Node* removeDuplicates(Node* head) {
    Node* current = head;
    while (current != nullptr) {
        Node* runner = current;
        while (runner->next != nullptr) {
            if (runner->next->data == current->data) {
                Node* temp = runner->next;
                runner->next = runner->next->next;
                delete temp;
            } else {
                runner = runner->next;
            }
        }
        current = current->next;
    }
    return head;
}

int main() {
    Node* head = createLinkedList();
    head = removeDuplicates(head);
    printLinkedList(head);
    return 0;
}