我正在嘗試找出一種方法來過濾掉陣列中的頂級分層元素，其中每個物件可能與陣列的其他元素具有父子層次結構。這是陣列的樣子：
[
{
"Code": "1A",
"name": "Western Europe",
"hierarchy": "World/",
},
{
"Code": "AT",
"name": "Austria",
"hierarchy": "World/ Western Europe/"
},
{
"Code": "NL",
"name": "Netherlands",
"hierarchy": "World/ Western Europe/"
}
]
對于此示例，輸出應該只是第一個物件，因為西歐是奧地利和荷蘭的父級。
[
{
"Code": "WO",
"name": "World",
"hierarchy": "",
},
{
"Code": "NA",
"name": "North America",
"hierarchy": "World/",
},
{
"Code": "NL",
"name": "Netherlands",
"hierarchy": "World/ Western Europe/"
}
]
這里的輸出將是 World，因為 World 是最高層次結構。
[
{
"Code": "1A",
"name": "Western Europe",
"hierarchy": "World/",
},
{
"Code": "NA",
"name": "North America",
"hierarchy": "World/",
},
{
"Code": "US",
"name": "United States",
"hierarchy": "World/ North America/",
},
{
"Code": "NL",
"name": "Netherlands",
"hierarchy": "World/ Western Europe/"
},
{
"Code": "RS",
"name": "Russia",
"hierarchy": "World/ Eastern Europe/",
},
]
在這里，輸出應該是西歐、北美和俄羅斯（因為我們在串列中沒有東歐，所以俄羅斯處于其層次結構的頂部）。所以，我們必須得到頂層的分層元素。我有一個函式可以傳遞代碼并獲取層次結構：
getParentChildHierarchy(string Code) { //這里有一些代碼 }
Input: AT
Output: WO>1A>AT
您能提出解決此問題的最佳方法嗎？
uj5u.com熱心網友回復：
一種方法是創建資料集中所有路徑的集合，即與 的連接hierarchy（name遵循與斜線和間距相同的格式）。然后檢查該集合中hierarchy實際存在哪些節點（參考父節點）。如果沒有，它在資料集中沒有父節點：
const data = [
{
"Code": "1A",
"name": "Western Europe",
"hierarchy": "World/",
}, {
"Code": "NA",
"name": "North America",
"hierarchy": "World/",
}, {
"Code": "US",
"name": "United States",
"hierarchy": "World/ North America/",
}, {
"Code": "NL",
"name": "Netherlands",
"hierarchy": "World/ Western Europe/"
}, {
"Code": "RS",
"name": "Russia",
"hierarchy": "World/ Eastern Europe/",
},
];
const keys = new Set(data.map(o => o.hierarchy " " o.name "/"));
const topLevel = data.filter(o => !keys.has(o.hierarchy));
console.log(topLevel);轉載請註明出處，本文鏈接：https://www.uj5u.com/net/520461.html
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