如何在本机运行leetcode链表题?

问题描述 投票:0回答:2

如何在本机运行链表程序? 当我在输入框中运行此代码时,它正在运行,但我似乎无法在本地计算机中运行此程序。

  function ListNode(val, next) {
      this.val = (val===undefined ? 0 : val)
      this.next = (next===undefined ? null : next)
  }
 
/**
 * @param {ListNode} list1
 * @param {ListNode} list2
 * @return {ListNode}
 */

var mergeTwoLists = function (l1, l2) {
  var mergedHead = { val: -1, next: null },
    crt = mergedHead;
  while (l1 && l2) {
    if (l1.val > l2.val) {
      crt.next = l2;
      l2 = l2.next;
    } else {
      crt.next = l1;
      l1 = l1.next;
    }
    crt = crt.next;
  }
  crt.next = l1 || l2;
  return mergedHead.next;
};

mergeTwoLists([1, 2, 4], [1, 3, 4]);
javascript linked-list
2个回答
3
投票

您可以使用这两个辅助函数将数组转换为链表,反之亦然,这本质上是 LeetCode 框架在幕后为您所做的事情。

const listFromArray = a => a.length ? new ListNode(a[0], listFromArray(a.slice(1)))  
                                    : null;
const arrayFromList = head => head ? [head.val].concat(arrayFromList(head.next)) 
                                   : [];

在您的情况下,您可以像这样使用它们:

const result = arrayFromList(
    mergeTwoLists(listFromArray([1, 2, 4]), listFromArray([1, 3, 4]))
);

0
投票

我能够通过提供我生成的

mergeTwoLists
ListNode
类在本地运行
LinkedList
方法。如下所示,
runCase
辅助函数采用两个
ListNodes
LinkedList
头),将它们合并并打印结果。我们可以使用
toArray
和 LeetCode 测试用例输入创建一个链表,并使用
fromArray
LinkedList
函数将结果转换为数组。

class ListNode {
  constructor(val=0, next=null) {
      this.val = val;
      this.next = next;             
  }
}

class LinkedList {
  constructor(head = null) {
    this.head = head
  }

  static fromArray(array) {
    const linkedList = new LinkedList();

    for (let i = array.length - 1; i >= 0; i--) {
      const node = new ListNode(array[i]);
      node.next = linkedList.head;
      linkedList.head = node;
    }

    return linkedList;
  }

  toArray() {
    const array = [];
    let current = this.head;

    while (current) {
      array.push(current.val);
      current = current.next;
    }

    return array;
  }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 *
 * @param {ListNode} node1
 * @param {ListNode} node2
 * @return {ListNode}
 */
var mergeTwoLists = function(node1, node2) {
  let linkedList = new LinkedList(),
  current = linkedList;

  while (node1 && node2) {
    if (node1.val > node2.val) {
      current.next = node2;
      node2 = node2.next;
    } else {
      current.next = node1;
      node1 = node1.next;
    }
    current = current.next;
  }
  current.next = node1 || node2;

  return linkedList.next;
};

const runCase = (array1, array2) => {
  const node1 = LinkedList.fromArray(array1).head;
  const node2 = LinkedList.fromArray(array2).head

  const result = mergeTwoLists(node1, node2);
  const linkedList = new LinkedList(result);
  console.log(linkedList.toArray());
}

runCase([1, 2, 4], [1, 3, 4]); // [1,1,2,3,4,4]
runCase([], []); // []
runCase([], [0]); // [0]

参考

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