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leetcode707 - Design Linked List - medium

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Design your implementation of the linked list. You can choose to use a singly or doubly linked list.
A node in a singly linked list should have two attributes: val and nextval is 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 the MyLinkedList class:

  • MyLinkedList() Initializes the MyLinkedList object.
  • int get(int index) Get the value of the indexth node in the linked list. If the index is invalid, return -1.
  • void addAtHead(int 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.
  • void addAtTail(int val) Append a node of value val as the last element of the linked list.
  • void addAtIndex(int index, int val) Add a node of value val before the indexth node in the linked list. If index equals the length of the linked list, the node will be appended to the end of the linked list. If index is greater than the length, the node will not be inserted.
  • void deleteAtIndex(int index) Delete the indexth node in the linked list, if the index is valid.

 

Example 1:

Input
["MyLinkedList", "addAtHead", "addAtTail", "addAtIndex", "get", "deleteAtIndex", "get"]
[[], [1], [3], [1, 2], [1], [1], [1]]
Output
[null, null, null, null, 2, null, 3]

Explanation
MyLinkedList myLinkedList = new MyLinkedList();
myLinkedList.addAtHead(1);
myLinkedList.addAtTail(3);
myLinkedList.addAtIndex(1, 2);    // linked list becomes 1->2->3
myLinkedList.get(1);              // return 2
myLinkedList.deleteAtIndex(1);    // now the linked list is 1->3
myLinkedList.get(1);              // return 3

 

Constraints:

  • 0 <= index, val <= 1000
  • Please do not use the built-in LinkedList library.
  • At most 2000 calls will be made to getaddAtHeadaddAtTail,  addAtIndex and deleteAtIndex.
 
练手题。注意细节:1. addAtTail的时候如果没head直接call addAtHead
2. traverse到tail是 while(p&&p->next) break出去后p指向tail
3. add/deleteAtIndex都要特殊考虑index为0的情况,traverse的时候记得希望p停在指向index的previous,这样p->next就刚好是要增/删的那个位置
4. delete刚好delete tail的话也是特殊考虑,直接把next设成nullptr
 
实现:
class MyLinkedList {
public:
    struct ListNode{
        ListNode* next;
        int val;
        ListNode(int x) : val(x), next(nullptr) {}
    };
    
    ListNode* head;
    /** Initialize your data structure here. */
    MyLinkedList() {
        head = nullptr;
    }
    
    /** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
    int get(int index) {
        if (index < 0)
            return -1;
        if (!head)
            return -1;
        ListNode* p = head;
        for (int i=0; i<index; i++){
            if (p->next)
                p = p->next;
            else
                return -1;
        }
        return p->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. */
    void addAtHead(int val) {
        ListNode* newNode = new ListNode(val);
        newNode->next = head;
        head = newNode;
    }
    
    /** Append a node of value val to the last element of the linked list. */
    void addAtTail(int val) {
        if (!head){
            addAtHead(val);
            return;
        }
        ListNode* newNode = new ListNode(val);
        ListNode* p = head;
        while (p && p->next){
            p = p->next;
        }
        p->next = newNode;
    }
    
    /** 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. */
    void addAtIndex(int index, int val) {
        if (index < 0) return;
        if (index == 0){
            addAtHead(val);
            return;
        }
        ListNode* newNode = new ListNode(val);
        ListNode* p = head;
        for (int i=0; i<index-1; i++){
            if (p->next) 
                p = p->next;
            else return;
        }
        ListNode* next = p->next;
        p->next = newNode;
        newNode->next = next;
    }
    
    /** Delete the index-th node in the linked list, if the index is valid. */
    void deleteAtIndex(int index) {
        if (index < 0) return;
        if (index == 0){
            head = head->next;
            return;
        }
        ListNode* p = head;
        for (int i=0; i<index-1; i++){
            if (p->next)
                p = p->next;
            else return;
        }
        if (p->next){
            ListNode* next = p->next->next;
            p->next = next;
        }
        else p->next = nullptr;
    }
};

 

leetcode707 - Design Linked List - medium

原文:https://www.cnblogs.com/xuningwang/p/13873022.html

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