setDataStructure.spec.js

import {expect, test} from 'vitest';
import {getDataStructure} from "#common/utils/question-handler/parseStructure.js";
import {showLogs} from "#common/utils/question-handler/showLogs.js";
import {ListNode} from "#common/structures/ListNode";

const jsDoc = `/**
* Definition for singly-linked list.
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode[]} lists
* @return {ListNode}
*/
var mergeKLists = function(lists) {
 // lists = convertListNode(lists);
 console.log(lists, '111')
// 当是空数组的情况下
if (!lists.length) {
 return null;
}
// 合并两个排序链表
const merge = (head1, head2) => {
 let dummy = new ListNode(0);
 let cur = dummy;
 // 新链表，新的值小就先接谁
 while (head1 && head2) {
   if (head1.val < head2.val) {
     cur.next = head1;
     head1 = head1.next;
   } else {
     cur.next = head2;
     head2 = head2.next;
   }
   cur = cur.next;
 }
 // 如果后面还有剩余的就把剩余的接上
 cur.next = head1 == null ? head2 : head1;
 return dummy.next;
};
const mergeLists = (lists, start, end) => {
 if (start + 1 == end) {
   return lists[start];
 }
 // 输入的k个排序链表，可以分成两部分，前k/2个链表和后k/2个链表
 // 如果将这前k/2个链表和后k/2个链表分别合并成两个排序的链表，再将两个排序的链表合并，那么所有链表都合并了
 let mid = (start + end) >> 1;
 let head1 = mergeLists(lists, start, mid);
 let head2 = mergeLists(lists, mid, end);
 return merge(head1, head2);
};
return mergeLists(lists, 0, lists.length);
};`
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function(lists) {
  // lists = convertListNode(lists);
  console.log(lists, '111')
// 当是空数组的情况下
if (!lists.length) {
  return null;
}
// 合并两个排序链表
const merge = (head1, head2) => {
  let dummy = new ListNode(0);
  let cur = dummy;
  // 新链表，新的值小就先接谁
  while (head1 && head2) {
    if (head1.val < head2.val) {
      cur.next = head1;
      head1 = head1.next;
    } else {
      cur.next = head2;
      head2 = head2.next;
    }
    cur = cur.next;
  }
  // 如果后面还有剩余的就把剩余的接上
  cur.next = head1 == null ? head2 : head1;
  return dummy.next;
};
const mergeLists = (lists, start, end) => {
  if (start + 1 == end) {
    return lists[start];
  }
  // 输入的k个排序链表，可以分成两部分，前k/2个链表和后k/2个链表
  // 如果将这前k/2个链表和后k/2个链表分别合并成两个排序的链表，再将两个排序的链表合并，那么所有链表都合并了
  let mid = (start + end) >> 1;
  let head1 = mergeLists(lists, start, mid);
  let head2 = mergeLists(lists, mid, end);
  return merge(head1, head2);
};
return mergeLists(lists, 0, lists.length);
};

test('测试', ()=> {

  const param = getDataStructure(jsDoc);
  const returnArray = getDataStructure(jsDoc, 'return');
  expect(param).toEqual([ 'ListNode[]'])
  expect(returnArray).toEqual(['ListNode'])

//   const listNode1 = [[[[1,4,5],[1,3,4],[2,6]]],[[]],[[[]]]];
// const listNode2 = [[1,1,2,3,4,4,5,6],[],[]];
//   showLogs(
//     mergeKLists,
//     listNode1,
//     listNode2
// )

})