Js 单链表 实践

js 实现单链表

// 节点
class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
  }
}

// 链表
class LinkedList {
  constructor() {
    this.head = null;
    this.tail = this.head; // 开始的时候, 指向的是同一个地址
    this.length = 0;
  }

  // 顺序插入
  append(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      this.tail = newNode;
    } else {
      this.tail.next = newNode;
      this.tail = newNode;
    }
    this.length++;
  }

  // 在链表头部插入
  prepend(value) {
    const newHead = new Node(value);
    newHead.next = this.head;
    this.head = newHead;
    this.length++;
  }

  // 在指定位置插入
  insert(value, index) {
    if (index >= this.length) {
      this.append(value);
    } else {
      const insetNode = new Node(value);
      const { prevNode, nextNode } = this.getPrevNextNode(index);
      prevNode.next = insetNode;
      insetNode.next = nextNode;
      this.length++;
    }
  }

  // 查找当前节点
  find(index) {
    let currNode = this.head;
    let flag = 0;
    while(flag < index - 1) {
      currNode = currNode.next;
      flag++;
    }
    return currNode;      
  }

  // 删除
  remove(index) {
    const { prevNode, nextNode } = this.getPrevNextNode(index);
    prevNode.next = nextNode;
  }
  
  // 反转
  reverse() {
    let previousNode = null;
    let currNode = this.head;
    while(currNode !== null) {
      let nextNode = currNode.next;
      currNode.next = previousNode;
      previousNode = currNode;
      currNode = nextNode;
    }
  }

  // 获取上一个和下一个结点
  getPrevNextNode(index) {
    let flag = 0;
    let prevNode = this.head;
    let nextNode = prevNode.next.next;
    while(flag < index - 1) {
      prevNode = prevNode.next;
      nextNode = prevNode.next;
      flag++;
    }
    return { prevNode, nextNode }
  }
}

const linkedList = new LinkedList();
linkedList.append(1);
linkedList.append(2);
linkedList.append(3);
linkedList.append(4);
linkedList.prepend(0);
console.log(linkedList.find(3))

LeetCode 链表相关的题型

一、链表的反转

1. 反转链表：leetcode.cn/problems/UH…
   输入:[1, 2, 3, 4, 5] 输出:[5, 4, 3, 2, 1]

    /**
     * @param {ListNode} head
    * @return {ListNode}
    */
    var reverseList = function(head) {
      let prev = null;
      let curr = head;
      while(curr) {
        const next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
      }
      return prev;
    };
   

2. 回文字符串：leetcode.cn/problems/aM… 输入: head = [1,2,3,3,2,1] 输出: true

   /**
     * Definition for singly-linked list.
    * function ListNode(val, next) {
    *     this.val = (val===undefined ? 0 : val)
    *     this.next = (next===undefined ? null : next)
    * }
    */
    /**
    * @param {ListNode} head
    * @return {boolean}
    */
    var isPalindrome = function(head) {
        let arr = [];
        while (head !== null) {
            arr.push(head.val)
            head = head.next;
        }
        let res = true;
        for (let i = 0, j = arr.length - 1; i < j; ++i, --j) {
            if (arr[i] !== arr[j]) {
                return false;
            }
        }
        return res;
    };