707. Design Linked List

🟠   🔖  设计 链表  🔗 LeetCode

题目

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 next. val 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 get, addAtHead, addAtTail, addAtIndex and deleteAtIndex.

题目大意

你可以选择使用单链表或者双链表，设计并实现自己的链表。

单链表中的节点应该具备两个属性：val 和 next 。val 是当前节点的值，next 是指向下一个节点的指针/引用。

如果是双向链表，则还需要属性 prev 以指示链表中的上一个节点。假设链表中的所有节点下标从 0 开始。

实现 MyLinkedList 类：

• MyLinkedList() 初始化 MyLinkedList 对象。
• int get(int index) 获取链表中下标为 index 的节点的值。如果下标无效，则返回 -1 。
• void addAtHead(int val) 将一个值为 val 的节点插入到链表中第一个元素之前。在插入完成后，新节点会成为链表的第一个节点。
• void addAtTail(int val) 将一个值为 val 的节点追加到链表中作为链表的最后一个元素。
• void addAtIndex(int index, int val) 将一个值为 val 的节点插入到链表中下标为 index 的节点之前。如果 index 等于链表的长度，那么该节点会被追加到链表的末尾。如果 index 比长度更大，该节点将 不会插入 到链表中。
• void deleteAtIndex(int index) 如果下标有效，则删除链表中下标为 index 的节点。

解题思路

链表的实现，详细可以参考 2.2 链表 - 链表的实现

代码

var MyLinkedList = function () {
  this.head = null;
};

var Node = function (val) {
  this.val = val;
  this.next = null;
};
/**
 * @param {number} index
 * @return {number}
 */
MyLinkedList.prototype.get = function (index) {
  if (index < 0 || index > 1000) return -1;
  let i = 0;
  let prev = this.head;
  while (prev) {
    if (i === index) break;
    prev = prev.next;
    i++;
  }
  return prev ? prev.val : -1;
};

/**
 * @param {number} val
 * @return {void}
 */
MyLinkedList.prototype.addAtHead = function (val) {
  const node = new Node(val);
  node.next = this.head;
  this.head = node;
};

/**
 * @param {number} val
 * @return {void}
 */
MyLinkedList.prototype.addAtTail = function (val) {
  const node = new Node(val);
  if (!this.head) {
    this.head = node;
  } else {
    let prev = this.head;
    while (prev.next) {
      prev = prev.next;
    }
    prev.next = node;
  }
};

/**
 * @param {number} index
 * @param {number} val
 * @return {void}
 */
MyLinkedList.prototype.addAtIndex = function (index, val) {
  if (index < 0 || index > 1000) return -1;
  const node = new Node(val);
  if (index === 0) {
    node.next = this.head;
    this.head = node;
  } else {
    let i = 1;
    let prev = this.head;
    while (prev) {
      if (i === index) {
        let temp = prev.next;
        prev.next = node;
        node.next = temp;
        break;
      }
      i++;
      prev = prev.next;
    }
  }
};

/**
 * @param {number} index
 * @return {void}
 */
MyLinkedList.prototype.deleteAtIndex = function (index) {
  if (index < 0 || index > 1000) return -1;
  if (index === 0) {
    this.head = this.head.next;
  } else {
    let i = 1;
    let prev = this.head;
    while (prev.next) {
      if (i === index) {
        let temp = prev.next;
        prev.next = temp.next;
        break;
      }
      i++;
      prev = prev.next;
    }
  }
};

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * var obj = new MyLinkedList()
 * var param_1 = obj.get(index)
 * obj.addAtHead(val)
 * obj.addAtTail(val)
 * obj.addAtIndex(index,val)
 * obj.deleteAtIndex(index)
 */

相关题目

相关题目

• 1206. 设计跳表

- [1206. 设计跳表](https://leetcode.com/problems/design-skiplist)