• Design Linked List


    Design Linked List
    
    Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. valis the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.
    
    Implement these functions in your linked list class:
    
    get(index) : Get the value of the index -th node in the linked list. If the index is invalid, return -1.
    addAtHead(val) : Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
    addAtTail(val) : Append a node of value val to the last element of the linked list.
    addAtIndex(index, val) : Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
    deleteAtIndex(index) : Delete theindex-th node in the linked list, if the index is valid.
    Example:
    
    MyLinkedList linkedList = new MyLinkedList();
    linkedList.addAtHead(1);
    linkedList.addAtTail(3);
    linkedList.addAtIndex(1, 2);  // linked list becomes 1->2->3
    linkedList.get(1);            // returns 2
    linkedList.deleteAtIndex(1);  // now the linked list is 1->3
    linkedList.get(1);            // returns 3
    Note:
    
    All values will be in the range of [1, 1000].
    The number of operations will be in the range of [1, 1000].
    Please do not use the built-in LinkedList library.
    思路:本题要求自己设计实现链表的基本操作,不能使用已有的模板函数,用双链表或者单链表均可。单链表处理起来稍微简单点,这里使用单链表进行操作,主要注意各个操作处理的细节,开始提交时忘记判断 index 小于 0 的情况,导致没有通过,加上这个判断就好了。
    
    
    #include <iostream>
    
    class Node{
    public:
        int val;
        Node* next;
        Node(int val) {this->val = val; next = nullptr;}
    };
    
    class MyLinkedList{
    public:
        int size = 0;
        Node* head = new Node(0);
        MyLinkedList(){}
    
        int get(int index)
        {
            if(index >= size || index < 0)
                return -1;
    
            Node* temp = head->next;
            for(int i = 0; i < index; i++)
                temp = temp->next;
    
            return temp->val;
        }
    
        void addAtHead(int val){
            Node* temp = head->next;
            head->next = new Node(val);
            head->next->next = temp;
            size++;
        }
    
        void addAtTail(int val){
            Node* temp = head;
            while (temp->next != nullptr) {
                temp = temp->next;
            }
            temp->next = new Node(val);
            size++;
        }
    
        void addAtIndex(int index, int val)
        {
            if(index > size)
                return;
    
            Node* temp = head;
            for(int i = 0; i < index; i++)
                temp = temp->next;
            Node* temp1 = temp->next;
            temp->next = new Node(val);
            temp->next->next = temp1;
            size++;
        }
    
        void deleteAtIndex(int index){
            if(index >= size || index < 0)
                return;
    
            Node* temp = head;
            for(int i = 0; i < index; i++)
                temp = temp->next;
    
            Node* temp1 = temp->next;
            temp->next = temp1->next;
            temp1->next = nullptr;
            size--;
            delete temp1;
        }
    };
    怕什么真理无穷,进一寸有一寸的欢喜。---胡适
  • 相关阅读:
    一周内签到连续天数求解
    int型动态数组总结
    快速排序总结
    希尔排序总结
    冒泡排序的总结
    桶排序总结
    插入排序的总结
    选择排序的总结
    二分法的五种实现
    安装Yum源
  • 原文地址:https://www.cnblogs.com/hujianglang/p/12192333.html
Copyright © 2020-2023  润新知