인프런 커뮤니티 질문&답변

prev와 tail을 이용해서 만들어 봤습니다!

궁금한점이 하나 있는데 remove 메서드의 if (current)의 else 부분은 필요하지 않은것 같아서 구현하지 않았는데 문제가 있지는 않나요?

class LinkedList {
  length = 0;
  head = null;
  tail = null;

  add(value) {
    const newNode = new Node(value);
    if (this.head) {
      this.tail.next = newNode;
      this.tail = newNode;
    } else {
      this.head = newNode;
      this.tail = newNode;
    }
    this.length++;
    return this.length;
  }
  search(index) {
    return this.#search(index)[1]?.value;
  }

  prevSearch(index) {
    return this.#search(index)[0]?.value;
  }
  #search(index) {
    let count = 0;
    let prev;
    let current = this.head;
    while(count < index) {
      prev = current;
      current = current?.next;
      count++;
    }
    return [prev, current];
  }
  remove(index) {
    const [prev, current] = this.#search(index);

    if (current) {
      if (prev) {
        prev.next = current.next;
      }
      if (current.next) {
        current.next.prev = prev;
      }
      if (current === this.tail) {
        this.tail = prev;
      }
    } 
    this.length--;
    return this.length;
  }
}

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

const li = new LinkedList();

li.add(1);
li.add(2);
li.add(3);
li.add(4);
li.add(5);
li.add(6);

console.log(li.prevSearch(2));
console.log(li.remove(4)); 
console.log(li.search(4)); 
console.log(li.tail.value);
console.log(li.remove(3));
console.log(li.tail.value);
console.log(li.remove(3));
console.log(li.tail.value);
console.log(li.remove(2));
console.log(li.remove(1));
console.log(li.tail.value);
console.log(li.remove(0));