我正在尝试从链接列表中删除重复项，但是重复项仍然存在

Question

我有以下类 ListNodeclass ListNode {    constructor(val) {        this.val = val        this.next = null    }}我试图通过将值存储在哈希中来删除重复值，如果存在这些值，则不应使用这些值形成新节点；但是，我所完成的只是当前列表的副本。我不确定我在哈希中做错了什么。我的逻辑结构不正确吗？const deleteDuplicates = (head) => {    let newNode = new ListNode('dummy')    let current = newNode    let headHash = {}    while (head) {        if (headHash[head.val] === undefined) {            if (newNode.next === null) newNode.next = new ListNode(head.val)            else {                newNode = newNode.next                newNode.next = new ListNode(head.val)            }        }        headHash = (headHash[head.val] || 0) + 1        console.log(headHash, head.val);        head = head.next    }    return current.next}const l = new ListNode(1)l.next = new ListNode(1)l.next.next = new ListNode(2)const l2 = new ListNode(1)l2.next = new ListNode(1)l2.next.next = new ListNode(2)l2.next.next.next = new ListNode(3)l2.next.next.next.next = new ListNode(3)console.log(deleteDuplicates(l));console.log(deleteDuplicates(l2));

Answer 1

违规行似乎是这一行：

headHash = (headHash[head.val] || 0) + 1

您不必将整体分配headHash给一个值，而是必须将headHash的当前头值属性分配给一个未定义的值。请看下面的代码：

class ListNode {

  constructor(val) {

    this.val = val;

    this.next = null;

  }

}

const deleteDuplicates = (head) => {

  let newNode = new ListNode("dummy");

  let current = newNode;

  let headHash = {};

  while (head) {

    if (headHash[head.val] === undefined) {

      if (newNode.next === null) {

          newNode.next = new ListNode(head.val);

      } else {

          newNode = newNode.next;

          newNode.next = new ListNode(head.val);

        }

    }

    // headHash = (headHash[head.val] || 0) + 1 // <- offending line

    headHash[head.val] = (headHash[head.val] || 0) + 1; // You need to assign the value to the headHash's key, not to the headHasn itself.

    head = head.next;

  }

  return current.next;

};

const l = new ListNode(1);

l.next = new ListNode(1);

l.next.next = new ListNode(2);

const l2 = new ListNode(1);

l2.next = new ListNode(1);

l2.next.next = new ListNode(2);

l2.next.next.next = new ListNode(3);

l2.next.next.next.next = new ListNode(3);

console.log(deleteDuplicates(l));

console.log(deleteDuplicates(l2));