const getSelectedSubTree = (nodes) => { const dummyRoot = { children: [] }; const dfs = (vertex, origin) => { if (vertex && vertex.isSelect) { if (!origin.children) origin.children = []; const newVertex = { ...vertex, children: [] }; origin.children.push(newVertex); if (!vertex.children) return; for (let i = 0; i < vertex.children.length; ++i) { dfs(vertex.children[i], newVertex); } } }; for (let i = 0; i < nodes.length; ++i) { dfs(nodes[i], dummyRoot); } return dummyRoot.children;};console.log(JSON.stringify(getSelectedSubTree(temp), null, 2));

也沒啥好說的，遞歸唄。findArrWebNameEq = function(arr){ var tag = false; var hash = {}; arr.forEach((item)=>{ if(hash[item.webName]) tag = true; hash[item.webName] = 1; if(item.childList){ if( findArrWebNameEq(item.childList)){ tag = true; } } }) return tag}

遍歷第 1 個數(shù)組，生成 key -> name 的映射表 遍歷第 2 個，查表替換 name

我不知道你樹形結(jié)構(gòu)是啥樣的，我這里用DeptTree做樹形結(jié)構(gòu)，你的那個{deptId,nextId}結(jié)構(gòu)的數(shù)據(jù)，我用Dept對應(yīng)。我也不確定你對遍歷的順序是否有要求，按照深度優(yōu)先算法寫了個。public class Sample { public static void main(String[] args) { DeptTree root = new DeptTree(1); DeptTree root2 = new DeptTree(2); DeptTree root5 = new DeptTree(5); root.addChild(root2); root.addChild(new DeptTree(3)); root.addChild(new DeptTree(4)); root2.addChild(root5); root2.addChild(new DeptTree(6)); root5.addChild(new DeptTree(7)); root5.addChild(new DeptTree(8)); List<Dept> result = new LinkedList<>(); traverse(root, result); System.out.println(result); } /** * 深度優(yōu)先遍歷 遍歷樹形list */ p

樹形轉(zhuǎn)換成數(shù)組數(shù)據(jù)let data = [ { "label": "城陽區(qū)", "children": [ { "label": "錦繡小學(xué)", "children": [ { "label": "1", "num": 1 }, { "label": "1、2", "num": 1 }, { "label": "2", "num": 2 }, { "label": "3", "num": 4 }, { "label": "3、4", "num": 2 }, { "label": "4", "num": 1 }, { "label": "4、5", "num": 1 }, { "label": "5", "num": 5 } ] }, { "label": "城陽區(qū)夏莊街道夏莊小學(xué)", "children": [ { "label": "無", "num": 13 } ] }, { "label": "夏莊街道營村小學(xué)", "children": [ { "label": "1", "num": 1 }, { "label": "1-2", "num": 3 }, { "label": "2", "num": 2 } ] }, { "label": "程哥莊小學(xué)", "children": [ { "label": "1", "num": 5 }, { "label": "2", "

本來說建議閱讀：從列表生成樹，但是文章里好像不是處理的這種結(jié)構(gòu)的列表。下面是對這個問題的答案function makeTree(data, keys) { return data.reduce((acc, it) => { let parent = { children: acc }; keys.map(key => it[key]) .forEach(value => { const children = (parent.children ??= []); parent = children.find(({ label }) => value === label) ?? children[children.push({ label: value }) - 1]; }); return acc; }, []);}const keys = Array.from(Array(3), (_, i) => `Zone${i + 1}`);const result = makeTree(testArray, keys);console.dir(result, { depth: null });思路和 @ForkKILLET 的差不多。里面那個循環(huán)本身也是一個遞進的過程，也可以寫成 reduce 的function makeTree(data, keys) { r

在線體驗地址const container = document.createElement("div");const treeNodeTemplate = (key, label) => `<div id="${key}"> <label>${label}</label> <input type="checkbox" /> <ul class="children"></ul> </div> `;const createTreeNodeFromKeyAndLabel = (key, label) => { const tmeplate = treeNodeTemplate(key, label); container.innerHTML = tmeplate; return container.children[0];};const effects = [];const visited = new Set();let flushed = false;class TreeNode { elem = null; parent = null; children = []; data = null; effects = []; constructor(data) { this.data = da

function solution(list, target) { let r ;(function findNode(list, target) { for (let item of list) { if (item.id === target) { r = item return item } if (Array.isArray(item.children)) { findNode(item.children, target) } } })(list, target) return r}const case1 = solution(originData, 1)const case2 = solution(originData, 111)const case3 = solution(originData, 3111)console.log('case1: ', case1);console.log('case2: ', case2);console.log('case3: ', case3);

遞歸，這種問題第一直覺一般就是遞歸var data = [{ name: '桑椹', id: '114' }, { name: '黑莓', id: '115' }, { name: '柑橘類', id: '116', children: [{ name: '橘子', id: '117' }, { name: '橙子', id: '118' }] }, { name: '奧特曼大全', id: '119', children: [{ name: '昭和類', id: '120', children: [{ name: '杰克', id: '121' }, { name: '泰羅', id: '122' }, { name: '艾斯', id: '123' }] }, { name: '平成類', id: '124', children: [{ name: '迪迦', id: '125' }, { name: '蓋亞', id: '126' }] }] }]function mapLastChild(list) { return list.reduce((res, item) => { if(item.children) res = res.concat(mapLastChild(item.children)) else res.push(item); return res; }

var data = [ { id: 1, name: "hei", children: [ { id: 3, name: "bai" }, { id: 4, name: "hong", children: [{ id: 6, name: "lv" }] } ] }, { id: 2, name: "jack", children: [{ id: 5, name: "lan" }] }];var result = [];function flatTree(data){ data.forEach(item => { result.push({ id: item.id, name: item.name }) if(item.children?.length){ result=flatTree(item.children); } }); return result}flatTree(data)

下列代碼可格式化 百度省市三級地域代碼：function format(data, parent = 0) { return data && Object.entries(data).map(([key, val]) => ({ id: val?.id ?? val, name: key, parent, children: format(val['市'] ?? val['區(qū)縣'], val.id) })) || null}const formatted_data = format(data);@JenK 我在這回復(fù)吧，評論里貼不了表格，也貼不了圖片，怪難受的。。感謝大佬的耐心回復(fù)。還是有個疑問：為啥不用些諸如 SQLite 的數(shù)據(jù)庫，替代 json 存儲 和 js 查詢呢？昨天，我測試了下，用 SQLite 來實現(xiàn)大佬的【地址智能識別】庫功能（主要是識別省/市/區(qū)/街/村，目前支持逐級模糊匹配），并利用 sql.js 在瀏覽器端測試（硬件：i5-8250U 筆記本）。結(jié)果： 行政區(qū)劃層級數(shù) 數(shù)據(jù)條數(shù) 查詢 7 個平均 39 字 4~5 級地址總耗時 數(shù)據(jù)庫大小 gzip 后 zstd 后 5 級 661824 40 毫秒 16.8 MB 5.72 MB 4.32 MB 4 級 44918 5 毫秒 880 KB 409 KB 330 KB 而且內(nèi)存占用幾乎就

謝邀 按照題主直觀的問題需求來看，就是想要把List<A>通過stream的groupingBy怎么轉(zhuǎn)換成List<B>，我可以給一個肯定答復(fù)。。。 當(dāng)然是不可以鴨 因為只用groupingBy，它的返回，再怎么用幾個參數(shù)，都只能是一個map，肯定轉(zhuǎn)化不成不是個B，所以必須要在groupingBy外用其他方式，比如像樓上的兄弟就是用groupingBy之后，生成的map再用entrySet進行循環(huán)，如果把樓上兄弟的構(gòu)造方法寫到B類里，估計看起來會好看點 public class B { private String name; private List<String> values; public B(Map.Entry<String, List<String>> entry){ this.name = entry.getKey(); this.values = entry.getValue(); }} 這樣的話，樓上兄弟的代碼就可以這么寫了 List<B> bList = list.stream() .collect(Collectors.groupingBy(A::getName, Collectors.mapping(A::getValue, Collectors.toList()))) .entrySe

var test = [ { id: 1, domain: '一級域1', children: [ { id: 11, domain: '二級域11', parentid: 1, children: [ { id: 111, domain: '三級域1121', parentid: 11, children: [{ id: 1111, parentid: 111, domain: '四級域111' }], }, { id: 111, domain: '三級域1311', parentid: 11, children: [{ id: 1111, parentid: 111, domain: '四級域111' }], }, { id: 111, domain: '三級域1411', parentid: 11, children: [{ id: 1111, parentid: 111, domain: '四級域111' }], }, { id: 111, domain: '三級域1511', parentid: 11, children: [{ id: 1111, parentid: 111, domain: '四級域111' }], }, { id: 111, domain: '三級域1611', parentid: 11, children: [{ id: 1111, parenti

你之前不是問過類似問題么js/vue/react 如何優(yōu)化以下代碼，參考那個問題的回答稍稍修改下：function filter(data) { function cb(node) { if(!node.children) { return true; } const children = node.children.filter(cb); return children.length === 0 ? false : (node.children = children); } return cb({ children: data });}// TESTfilter(data)

function disabledTree(arr) { arr.forEach((el) => { el.isDisabled = el.hasChildren; disabledTree(el.children || []); });}let aa = [{ "children": [{ "children": [{ "children": [{ "hasChildren": false, "name": "小苔便利店", "pid": 587, "id": 1428553074498211840, "label": "小苔便利店", "deptType": 2, "leaf": true }], "hasChildren": true, "name": "天橋區(qū)", "pid": 586, "id": 587, "label": "天橋區(qū)", "deptType": 1, "leaf": false }, { "hasChildren": false, "name": "市中區(qū)", "pid": 586, "id": 588, "label": "市中區(qū)", "deptType": 1, "leaf": true }], "hasChildren": true, "name": "濟南市", "pid": 585, "id": 586, "label": "濟南市",