Question

Hi Stack Overflow社区，

我正在为电子商务网站构建搜索功能，并寻找满足多项要求的最佳 分页算法。

当用户在我们的网站上提交搜索产品查询时，我们会返回根据其指定的偏好评分的产品列表。然后在几个网页上显示搜索结果。由于多个商家信息可能属于同一个商家，因此我们希望确保商家不会在单个页面中控制结果。

结果按照降序分数顺序排序。
根据每页结果的数量，我们将结果集划分为单独的页面（ paginate ），同时在每个页面上最多出现一次商家。
如果某个页面无法仅使用唯一商家填充，然后该页面将填充剩余的商品条目，同时保留初始订购

例如： //每页显示的结果让numPerPage = 4;

// results array of csv, comma-delimited strings with format:
// merchant_id,listing_id,score,description

let results = [
 "5,90,500.1,description text",  
 "4,51,495.8,description text",  
 "8,72,490,description text",  
 "5,89,485.4,description text",
 "5,80,480,description text",
 "9,18,478.1,description text",  
 "6,50,475.9,description text",  
 "5,56,472,description text",  
 "7,12,470.5,description text",  
 "5,5,465,description text",
 "6,10,450,description text"
]

// Paginate function output: array of strings representing paginated results, 
// with an empty String separator after each page, except the last page.
let output = [
 "5,90,500.1,description text",  
 "4,51,495.8,description text",  
 "8,72,490,description text", 
 "9,18,478.1,description text",
 "", // this is a page separator
 "5,89,485.4,description text",
 "6,50,475.9,description text",  
 "7,12,470.5,description text",
 "6,10,450,description text",
 "", // this is a page separator
 "5,80,480,description text",
 "5,56,472,description text",  
 "5,5,465,description text"]

到目前为止，已经编写了以下代码：

function paginate(numPerPage, results) {
 // first map : transform into array of arrays
 // second map : transform into array of objects w/ key-value pairs
 let hash = {};
 let output = [];
 output = results.map(listing => listing.split(',')).map((listing, i) => {
    return {
      "index": i,
      "merchant_id": listing[0],
      "listing_id": listing[1],
      "score": listing[2],
      "description": listing[3]
    }
  });
  // sort the listings by descending score.
  output = output.sort((a,b) => b.score-a.score);
  console.log(output);
 // next, create a hash of merchant: array of listings
 // while results per page < numPerPage ...
 // ... optimal pagination algorithm in progress..
 }

我正在考虑在我们前进时创建"merchant": [array of that merchant's listings]和shift()商家数组的第一个值的哈希值。但是，当考虑每页要求的唯一商家偏好时，该算法会变得有趣。另外，我提出的算法并不是最优的。

真的很感激任何帮助！

Answer 1

我认为这可以通过set和Map

的组合来解决

key - MerchantId, 
value - PriorityQueue of type MerchantListingDetails

此处优先级队列是基于分数的最大堆。

MerchantListingDetails类由以下组成 -
MERCHANT_ID， listing_id，得分了，描述

1. Initially your Set & Map is empty. 
2. for each MerchantListingDetails // from input line 
   check if merchantID is present in set, 
    if not then add it to set and append to page1.
    if merchantId present in set, then add the merchantId and corresponding object to Map.
3. Continue till your page is full.

Now for page2 :
1. Delete all the entries from set.
2. copy the keys(merchantId's) from Map to set.
3. for each entry from Map,
     get the first entry for each merchant and display it on page.
     if the merchant has no entries left in map , delete its key from map and set.
4. If the page has required list of output, then again go to step 1 for new page and continue.

注意：这只是我在10分钟内想到的粗略算法。让我知道你是如何实现的。

分页算法：显示独特电子商务商家的排名搜索结果

1 个答案: