一家公司让做 4 个面试题，投入不少精力做完之后，就回了句代码风格不规范，再问就没回应了。不知道具体哪里不规范应该怎样改进？

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sxxkgoogle 2020-04-10 22:04:59 +08:00 3710 次点击

这是一个创建于 2024 天前的主题，其中的信息可能已经有所发展或是发生改变。

一家公司让做 4 个面试题，投入不少精力做完之后，就回了句代码风格不规范，再问就没回应了。不知道具体哪里不规范应该怎样改进？贴出来题目和我的答案，请大家指点赐教。

第一题, TreeNode 查询已知有如下一个树状数据结构：

let tree = { id: '1', label: 'first', children: [ { id: '2', label: 'second' }, { id: '3', label: 'third', children: [ { id: '4', label: 'fourth' }, { id: '5', label: 'fifth' } ] } ] };

请实现一个查询函数，通过输入 Tree 的 Root Node 和 Id，返回与其匹配的节点，函数原型如下（注意：请不要在函数内部直接用 console.log 打印出来）：

findNodeById(root: TreeNode, id: string): TreeNode;

这是第一题我的答案（虽然题目不是 java 的，但他们邮件里提到可以用任意语言作答）：

public class GetTreeNode { public static void main(String[] args) { String json1 = "{\r\n" + " id: '1',\r\n" + " label: 'first',\r\n" + " children: [\r\n" + " {\r\n" + " id: '2',\r\n" + " label: 'second'\r\n" + " },\r\n" + " {\r\n" + " id: '3',\r\n" + " label: 'third',\r\n" + " children: [\r\n" + " {\r\n" + " id: '4',\r\n" + " label: 'fourth'\r\n" + " },\r\n" + " {\r\n" + " id: '5',\r\n" + " label: 'fifth'\r\n" + " }\r\n" + " ]\r\n" + " }\r\n" + " ]\r\n" + "}"; JSONObject jObject1 = new JSONObject(json1); String id = "2"; JSONObject returnNode = findNodeById(jObject1, id); System.out.println(returnNode); } private static JSONObject findNodeById(JSONObject jObject1, String id) { JSONObject returnNode = null; for (int i = 0; i < jObject1.names().length(); i ++) { String key = jObject1.names().getString(i); if (key.equals("id")) { String value = jObject1.getString(key); if (((String) value).equals(id)) { returnNode = new JSONObject(jObject1.toString()); return returnNode; } } else { Object value = jObject1.get(key); if (returnNode == null) { returnNode = handleValue(value, id); } else { handleValue(value, id); } } } return returnNode; } private static JSONObject handleValue(Object value, String id) { JSONObject returnNodeh = null; if (value instanceof JSONObject) { returnNodeh = findNodeById((JSONObject) value, id); } else if (value instanceof JSONArray) { returnNodeh = handleJSONArray((JSONArray) value, id); } return returnNodeh; } private static JSONObject handleJSONArray(JSONArray jsonArray, String id) { JSONObject returnNodea = null; for (int i = 0; i < jsonArray.length(); i ++) { if (returnNodea == null) { returnNodea = handleValue(jsonArray.get(i), id); } else { handleValue(jsonArray.get(i), id); } } return returnNodea; } }

第二题：学生按成绩分组实现一个 groupBy 函数，将学生按照成绩等级进行分组。

// 成绩等级分为 A 、B 和 C 三级 function getGrade(score){ return score < 60 ? 'C' : score < 80 ? 'B' : 'A'; }; // 学生及其成绩 let students = [ {name: '张三', score: 84}, {name: '李四', score: 58}, {name: '王五', score: 99}, {name: '赵六', score: 69} ];

实现该函数：groupBy(students); 输出为：

{ 'A': [ {name: '王五', score: 99}, {name: '张三', score: 84} ], 'B': [{name: '赵六', score: 69}], 'C': [{name: '李四', score: 58}] }

第二题答案：

public class GroupByStudents { public static void main(String[] args) { String input = "[\r\n" + "{name: '张三', score: 84},\r\n" + "{name: '李四', score: 58},\r\n" + "{name: '王五', score: 99},\r\n" + "{name: '赵六', score: 69}\r\n" + "];"; JSONArray students = new JSONArray(input); JSONObject studentGroups = groupBy(students); System.out.println(studentGroups); } private static String getGrade(int score) { return score < 60 ? "C" : score < 80 ? "B" : "A"; } private static JSONObject groupBy(JSONArray students) { JSONObject studentGroups = new JSONObject("{}"); for (int i = 0; i < students.length(); i ++) { JSONObject student = students.getJSONObject(i); int score = student.getInt("score"); String grade = getGrade(score); studentGroups.append(grade, student); } return studentGroups; } }

第三题：字符串 Parse 请设计一个字符串 parse 函数，可以将输入的字符串分解为对应的树状结构，比如：

// 例子 1 let input = 'int'; let result = parse(intput); // result 结果为： // { // type: 'int' // }; // 例子 2 let input = 'Array<bool>'; let result = parse(intput); // { // type: 'Array', // typeArgs: [{ // type: 'bool' // }] // }; // 例子 3 let input = 'Array<Array<string>>'; let result = parse(intput); // { // type: 'Array', // typeArgs: [{ // type: 'Array', // typeArgs: [{ // type: 'string' // }] // }] // }; // 例子 4 let input = 'Map<string, Array<bool>>'; let result = parse(intput); // { // type: 'Map', // typeArgs: [{ // type: 'string' // }, { // type: 'Array', // typeArgs: [{ // type: 'bool' // }] // }] // };

同理，该 parse 函数可以分解如下任意情况，比如：

"int" "string" "bool" "Array<int>" "Array<Array<int>>" "Array<Array<Array<int>>>" "Map<string, Map<string, bool>>" "Map<Map<string, bool>, bool>"

第三题答案：

public class ParseString { public static void main(String[] args) { String input = "Map<string, Map<string, bool>>"; JSONObject result = parse(input); System.out.println(result); } private static JSONObject parse(String input) { JSONObject jsOnObj= new JSONObject("{}"); String[] inputSplit = input.split("<", 2); jsonObj.append("type", inputSplit[0]); if (inputSplit.length == 2) { loopParse(inputSplit[1].substring(0, inputSplit[1].length() - 1), jsonObj); } return jsonObj; } private static void loopParse(String string, JSONObject jsonObj) { List<String> result = new ArrayList<String>(); Stack<String> stackInQuotes = new Stack<>(); int start = 0; boolean inQuotes = false; for (int current = 0; current < string.length(); current++) { if (string.charAt(current) == '<') { stackInQuotes.add("<"); } else if (string.charAt(current) == '>') { stackInQuotes.pop(); } boolean atLastChar = (current == string.length() - 1); if (atLastChar) { result.add(string.substring(start)); } else if (string.charAt(current) == ',' && stackInQuotes.isEmpty()) { result.add(string.substring(start, current)); start = current + 1; } } JSONArray substringJsOnArr= new JSONArray(); for (int i = 0; i < result.size(); i ++) { String substring = result.get(i); String[] substringSplit = substring.split("<", 2); JSONObject substringJsOnObj= new JSONObject(); substringJsonObj.append("type", substringSplit[0]); if (substringSplit.length == 2) { loopParse(substringSplit[1].substring(0, substringSplit[1].length() - 1), substringJsonObj); } substringJsonArr.put(substringJsonObj); } jsonObj.append("tyeArgs", substringJsonArr); } }

第四题：自增 ID 已知有如下一个树状数据结构：

let tree = { id: '1', type: 'View', name: 'view', children: [ { id: '2', type: 'Button', name: 'button' }, { id: '3', type: 'View', name: 'view_1', children: [ { id: '4', type: 'Button', name: 'button_1' }, { id: '5', type: 'View', name: 'view_2' } ] } ] };

name 字段的规则为整棵树不能重复，如果遇到重复，则添加 _n 作为结尾。如果因为删除某些节点，名字出现了空隙，比如 Tree 中有 button_1 和 button_3，但是没有 button_2，下一次插入时，需要优先使用 button_2，而不是 button_4 。请实现一个唯一名称算法，当向整棵树的任意 children 插入一个新的 Node 时，可以保证 name 不重复。 srcName 是计划向 Tree 中插入节点的 name，比如 button 、view, rootTreeNode 是整棵树

function getIncName(srcName: string, rootTreeNode : TreeNode): string;

第四题答案：

public class GetIncName { public static void main(String[] args) { String json1 = "{\r\n" + " id: '1',\r\n" + " type: 'View',\r\n" + " name: 'view',\r\n" + " children: [\r\n" + " {\r\n" + " id: '2',\r\n" + " type: 'Button',\r\n" + " name: 'button'\r\n" + " },\r\n" + " {\r\n" + " id: '3',\r\n" + " type: 'View',\r\n" + " name: 'view_1',\r\n" + " children: [\r\n" + " {\r\n" + " id: '4',\r\n" + " type: 'Button',\r\n" + " name: 'button_1'\r\n" + " },\r\n" + " {\r\n" + " id: '5',\r\n" + " type: 'View',\r\n" + " name: 'view_5'\r\n" + " }\r\n" + " ]\r\n" + " }\r\n" + " ]\r\n" + "}"; JSONObject jObject1 = new JSONObject(json1); String srcName = "view"; String string = getIncName(srcName, jObject1); System.out.println(string); } private static String getIncName(String srcName, JSONObject jObject1) { Map<String, ArrayList<Integer>> namesMap = new LinkedHashMap<String, ArrayList<Integer>>(); handleJSONObject(jObject1, namesMap); String incName = checkSrcName(srcName, namesMap); return incName; } private static void handleJSONObject(JSONObject jObject1, Map<String, ArrayList<Integer>> namesMap) { jObject1.keys().forEachRemaining(key -> { Object value = jObject1.get(key); if (key.equals("name")) { String[] nameSplit = ((String) value).split("_"); if (namesMap.containsKey(nameSplit[0])) { if (nameSplit.length == 2) { ArrayList<Integer> nameList = namesMap.get(nameSplit[0]); nameList.add(Integer.valueOf(Integer.parseInt(nameSplit[1]))); namesMap.put(nameSplit[0], nameList); } else { ArrayList<Integer> nameList = namesMap.get(nameSplit[0]); nameList.add(Integer.valueOf(Integer.parseInt("0"))); namesMap.put(nameSplit[0], nameList); } } else { if (nameSplit.length == 2) { ArrayList<Integer> newNameList = new ArrayList<Integer>(); newNameList.add(Integer.valueOf(Integer.parseInt(nameSplit[1]))); namesMap.put(nameSplit[0], newNameList); } else { ArrayList<Integer> newNameList = new ArrayList<Integer>(); newNameList.add(Integer.valueOf(Integer.parseInt("0"))); namesMap.put(nameSplit[0], newNameList); } } } handleValue(value, namesMap); }); } private static void handleValue(Object value, Map<String, ArrayList<Integer>> namesMap) { if (value instanceof JSONObject) { handleJSONObject((JSONObject) value, namesMap); } else if (value instanceof JSONArray) { handleJSONArray((JSONArray) value, namesMap); } } private static void handleJSONArray(JSONArray jsonArray, Map<String, ArrayList<Integer>> namesMap) { jsonArray.iterator().forEachRemaining(element -> { handleValue(element, namesMap); }); } private static String checkSrcName(String srcName, Map<String, ArrayList<Integer>> namesMap) { String[] nameSplit = srcName.split("_"); if (namesMap.containsKey(nameSplit[0])) { ArrayList<Integer> nameList = namesMap.get(nameSplit[0]); Collections.sort(nameList); int availablenum = 0; for (int i = 0; i < nameList.size(); i ++) { if (nameList.get(i).intValue() != i) { availablenum = i; break; } else { availablenum = i + 1; } } return nameSplit[0]+"_"+availablenum; } else { return srcName; } } }

25 条回复 2020-04-11 11:52:57 +08:00

mumbler

2020-04-10 22:09:38 +08:00 via Android

代码都没有注释啊

sxxkgoogle

2020-04-10 22:20:27 +08:00

@mumbler 嗯没注释，之前在网上搜好像说做面试题一般不需要吧。

&bsp; 3

also24

2020-04-10 22:25:36 +08:00

怎么感觉题目是 TypeScript 的，楼主你这面的是什么岗位？

sxxkgoogle

2020-04-10 22:39:26 +08:00

@also24 是全栈的岗位。

mcfog

2020-04-10 22:46:54 +08:00 via Android

你看你就这样贴了题目出来，还有猎头也会收集题目，培训班也有一个个安排面试拿题目给下一个的操作，所以一般不会解释过多的

raymanr

2020-04-10 22:53:49 +08:00

呃，我不是专业程序员，好奇自己的水平在专业人员眼里是啥样的，做了下第一题，各位大佬点评下

```

function findNodeById(root, id) {
let result = [];

function findOne(node, id) {
if (node["id"] == id) {
result.push(node);
}
if (node.hasOwnProperty("children")) {
node["children"].forEach(child => findOne(child, id));
}
return false;
}

findOne(root, id);
return result;
}

```

swulling

2020-04-10 22:54:07 +08:00 via iPad

代码长度相比于出题人的预期略长了

aureole999

2020-04-10 23:05:26 +08:00

为什么不管什么都用 JsonObject 做呢？
题目只说有个树状结构，例子给的是 js，不代表非要用 json 来实现树，你可以自己定义 TreeNode Class 啊。如果题目要求必须用 json，那也应该把 json 映射成 java bean 再写处理的方法吧。要不你就别用 java，直接用 js 。

aureole999

2020-04-10 23:13:49 +08:00

@raymanr 第一题返回值不需要用数组。一般来说叫函数名 byId 的返回值就是唯一的，而且人家给的函数原型的返回值也不是数组。

ASpiral

2020-04-11 00:26:11 +08:00

试着做了下第一题，感觉没必要写那么长的代码吧…
const findNodeById = (root, id) => {
let target = null;
const findNode = (root, id) => {
if (target !== null) {
return;
}
if (root.id === id) {
target = root;
} else if (root.children) {
root.children.forEach(node => findNode(node, id));
}
};
findNode(root, id);
return target;
};

lithbitren

2020-04-11 02:33:06 +08:00

代码没细看，java 选手好可怕，感觉 python 都是几行以内解决的，js 写起来也就是多了半对大括号的行数。。

rabbbit

2020-04-11 03:36:06 +08:00

1
interface TreeNode {
...id: String,
...label: String,
...children?: TreeNode[]
}

function findNodeById(root: TreeNode, id: string): TreeNode {
...if (root.id === id) {
......return root;
...}
...
...if (root.children) {
......for (let i of root.children) {
.........const child = findNodeById(i, id);
.........if (child) {
............return child
.........}
......}
...} else {
... return null;
...}
}

console.assert(findNodeById(tree, '1').label === 'first', '1')
console.assert(findNodeById(tree, '2').label === 'second', '2')
console.assert(findNodeById(tree, '3').label === 'third', '3')
console.assert(findNodeById(tree, '4').label === 'fourth', '4')
console.assert(findNodeById(tree, '5').label === 'fifth', '5')

rabbbit

2020-04-11 03:51:58 +08:00

2
interface Student {
...name: string,
...score: number
}
function groupBy(students: Student[]) {
...const group: { [propName: string]: Student[] } = {};
...for (let i of students) {
......const grade = getGrade(i.score);
......if (!group[grade]) {
.........group[grade] = []
......}
......group[grade].push(i)
...}
...return group;
}

6IbA2bj5ip3tK49j

2020-04-11 04:03:08 +08:00 via iPhone

1，不要硬编码数据，从文件读取。方便测试，修改。
2，lambda

反正我作为 Java 开发，这样的代码风格，我是接受不了的。

lithbitren

2020-04-11 04:35:31 +08:00

几行夸张了，手撕中等题也要十行左右了。

宽搜扩展完事，这里就用字典代替树了，如果树是对象的话就更改获取属性的语句就行。
def findNodeById(root, id):
d = root and [root]
while d:
r = next((r for r in d if r['id'] == id), None)
if r:
return r
d = sum((r['children'] for r in d if 'children' in r), [])
return None

第二题最简单
def getGrade(score):
return score < 60 and 'C' or score < 80 and 'B' or 'A'
def groupBy(students):
res = {'A': [], 'B': [], 'C': []} # res = collections.defaultdict(list)
for student in students:
res[getGrade(student['score'])].append(student)
return res

语法题栈实现
def parse(args):
stack = [{'type: ''}]
for c in args:
if c == ' ':
continue
elif c == '<':
stack[-1]['typeArgs'] = [{'type': ''}]
stack.append(stack[-1]['typeArgs'][0])
elif c == ',':
stack[-2]['typeArgs'].append({'type': ''})
stack[-1] = stack[-2]['typeArgs'][-1]
elif c == '>':
del stack[-1]
else:
stack[-1]['type'] += c
return stack[0]

O(N)的在线算法，深搜遍历，计数判断，如果离线的话可以优化到 O(logN)
def getIncName(srcName, rootTreeNode):
d = set()
def dfs(r):
r['name'].rsplit('_', 1)[0] == srcName and d.add(r['name'])
if 'children' in r:
for child in r['children']:
dfs(child)
rootTreeNode and dfs(rootTreeNode)
if srcName not in d:
return srcName
for i in range(1, len(d) + 1):
res = srcName + '_' + str(i)
if res not in d:
return res

rabbbit

2020-04-11 05:24:58 +08:00

@lithbitren 空格是怎么打出来的?

lithbitren

2020-04-11 05:34:27 +08:00

@rabbbit 随便找了个空白字符，ascii 码 12644

tyx1703

2020-04-11 05:37:11 +08:00 via iPhone

@ASpiral 你这把整个树都遍历了
@rabbbit 的答案比较好

tyx1703

2020-04-11 05:42:39 +08:00 via iPhone

题目用的 ts，你用 java 作答我寻思着起码手动把 TreeNode 的数据结构实现一下吧
一坨坨 json 感官上就很不舒服，更加无心阅读代码质量了

yazoox

2020-04-11 08:10:40 +08:00 via Android

有意思，关注一下

alphatoad

2020-04-11 08:11:25 +08:00

别想太多，可能只是因为面试有 kpi

easylee

2020-04-11 08:15:53 +08:00 via Android

别想太多，你能来发帖，大概率不是你的问题。

我面过一些公司，面试题竟然有线性代数，还有多积分的！
还有一些公司面试题很快的写完了，结果不给回复了。

yhxx

2020-04-11 09:27:09 +08:00

可能因为出题的是前端，看到你实现 json 的方式觉得有点丑？

wangbot1

2020-04-11 11:47:28 +08:00

Java 的还是定义类型比较好, JsonNode 看起来不直观

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.CollectionType;
import lombok.Data;
import lombok.NonNull;

import java.io.IOException;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class Main {

@Data
static class TreeNode < T extends TreeNode > {
private List < T > children;
}

@Data
static class Label extends TreeNode < Label > {
private String id;
private String label;
}

@Data
static class OrderNameTreeNode extends TreeNode < OrderNameTreeNode > {
private String id;
private String type;
private String name;
}

@Data
static class Student {
private String name;
private double score;
}

@Data
static class Type {
private String type = "";
private List < Type > typeArgs = new ArrayList < > ();
}

@FunctionalInterface
public interface Recursion < T > {
void go(T t, Recursion < T > self);
}

private static Label question1(@NonNull Label root, @NonNull String id) {
if (id.equals(root.getId())) {
return root;
} else if (null != root.getChildren()) {

for (Label child: root.getChildren()) {
Label tartget = question1(child, id);

if (tartget != null) {
return tartget;
}
}

}
return null;
}

private static Map < String, List < Student >> question2(@NonNull List < Student > students) {
return students
.stream()
.collect(Collectors.groupingBy(student - >
student.getScore() < 60 ? "C" : student.getScore() < 80 ? "B" : "A"));
}

private static Type question3(@NonNull String typeArgs) {
List < Type > types = new ArrayList < > ();
Type root = new Type();
types.add(root);
for (int i = 0, len = typeArgs.length(); i < len; i++) {
char c = typeArgs.charAt(i);
if (c == '<') {
Type type = new Type();
types.get(types.size() - 1).getTypeArgs().add(type);
types.add(type);
} else if (c == ',') {
types.remove(types.size() - 1);
Type type = new Type();
types.get(types.size() - 1).getTypeArgs().add(type);
types.add(type);
} else if (c == ' ') {
continue;
} else if (c == '>') {
types.remove(types.size() - 1);
} else {
Type type = types.get(types.size() - 1);
type.setType(type.getType() + c);
}
}
return root;
}

private static String question4(@NonNull String srcName, @NonNull OrderNameTreeNode root) {
List < Integer > existsIndexs = new ArrayList < > ();
Recursion < OrderNameTreeNode > func = (treeNode, self) - > {

if (treeNode.getName().startsWith(srcName)) {
String substring = treeNode.getName().substring(srcName.length());
if ("".equals(substring)) {
existsIndexs.add(0);
} else {
existsIndexs.add(Integer.parseInt(substring.substring(1)));
}
}

if (null != treeNode.getChildren()) {
treeNode.getChildren().forEach(child - > self.go(child, self));
}

};
func.go(root, func);

if (existsIndexs.isEmpty()) {
return srcName;
}

existsIndexs.sort(Comparator.comparingInt(o - > o));

if (existsIndexs.get(0) != 0) {
return srcName;
}

for (int i = 1; i < existsIndexs.size(); i++) {
if (existsIndexs.get(i) - existsIndexs.get(i - 1) > 1) {
return String.format("%s_%d", srcName, i + 1);
}
}

return String.format("%s_%d", srcName, existsIndexs.size());
}

public static void main(String[] args) throws IOException {
ObjectMapper mapper = new ObjectMapper();

String treeNodeJson =
"{\"id\":\"1\",\"label\":\"first\",\"children\":[{\"id\":\"2\",\"label\":\"second\"},{\"id\":\"3\",\"label\":\"third\",\"children\":[{\"id\":\"4\",\"label\":\"fourth\"},{\"id\":\"5\",\"label\":\"fifth\"}]}]}";
Label treeNode = mapper.readValue(treeNodeJson, Label.class);
for (int i = 1; i < 5; i++) {
System.out.println(question1(treeNode, String.valueOf(i)));
}

String studentJson =
"[{\"name\":\"张三\",\"score\":84},{\"name\":\"李四\",\"score\":58},{\"name\":\"王五\",\"score\":99},{\"name\":\"赵六\",\"score\":69}]";
CollectionType collectiOnType= mapper.getTypeFactory().constructCollectionType(List.class, Student.class);
List < Student > students = mapper.readValue(studentJson, collectionType);
System.out.printf(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question2(students)));

System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question3("int")));
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question3("bool")));
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question3("Array<int>")));
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question3("Array<Array<int>>")));
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question3(
"Array<Array<Array<int>>>")));
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question3(
"Map<string, Map<string, bool>>")));
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(question3(
"Map<Map<string, bool>, bool>")));

String orderNameTreeNodeJson =
"{\"id\":\"1\",\"type\":\"View\",\"name\":\"view\",\"children\":[{\"id\":\"2\",\"type\":\"Button\",\"name\":\"button\"},{\"id\":\"3\",\"type\":\"View\",\"name\":\"view_1\",\"children\":[{\"id\":\"4\",\"type\":\"Button\",\"name\":\"button_1\"},{\"id\":\"5\",\"type\":\"View\",\"name\":\"view_2\"}]}]}";
OrderNameTreeNode orderNameTreeNode = mapper.readValue(orderNameTreeNodeJson, OrderNameTreeNode.class);
System.out.println(question4("view", orderNameTreeNode));
System.out.println(question4("button", orderNameTreeNode));
}

}

wangbot1

2020-04-11 11:52:57 +08:00

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.CollectionType;
import lombok.Data;
import lombok.NonNull;

import java.io.IOException;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class Main {

......@Data
......static class TreeNode < T extends TreeNode > {
............private List < T > children;
......}

......@Data
......static class Label extends TreeNode < Label > {
............private String id;
............private String label;
......}

......@Data
......static class OrderNameTreeNode extends TreeNode < OrderNameTreeNode > {
............private String id;
............private String type;
............private String name;
......}

......@Data
......static class Student {
............private String name;
............private double score;
......}

......@Data
......static class Type {
............private String type = "";
............private List < Type > typeArgs = new ArrayList < > ();
......}

......@FunctionalInterface
......public interface Recursion < T > {
............void go(T t, Recursion < T > self);
......}

......private static Label question1(@NonNull Label root, @NonNull String id) {
............if (id.equals(root.getId())) {
..................return root;
............} else if (null != root.getChildren()) {

..................for (Label child: root.getChildren()) {
........................Label tartget = question1(child, id);

........................if (tartget != null) {
..............................return tartget;
........................}
..................}

............}
............return null;
......}

......private static Map < String, List < Student >> question2(@NonNull List < Student > students) {
............return students
...................stream()
...................collect(Collectors.groupingBy(student - >
........................student.getScore() < 60 ? "C" : student.getScore() < 80 ? "B" : "A"));
......}

......private static Type question3(@NonNull String typeArgs) {
............List < Type > types = new ArrayList < > ();
............Type root = new Type();
............types.add(root);
............for (int i = 0, len = typeArgs.length(); i < len; i++) {
..................char c = typeArgs.charAt(i);
..................if (c == '<') {
........................Type type = new Type();
........................types.get(types.size() - 1).getTypeArgs().add(type);
........................types.add(type);
..................} else if (c == ',') {
........................types.remove(types.size() - 1);
........................Type type = new Type();
........................types.get(types.size() - 1).getTypeArgs().add(type);
........................types.add(type);
..................} else if (c == ' ') {
........................continue;
..................} else if (c == '>') {
........................types.remove(types.size() - 1);
..................} else {
........................Type type = types.get(types.size() - 1);
........................type.setType(type.getType() + c);
..................}
............}
............return root;
......}

......private static String question4(@NonNull String srcName, @NonNull OrderNameTreeNode root) {
............List < Integer > existsIndexs = new ArrayList < > ();
............Recursion < OrderNameTreeNode > func = (treeNode, self) - > {

..................if (treeNode.getName().startsWith(srcName)) {
........................String substring = treeNode.getName().substring(srcName.length());
........................if ("".equals(substring)) {
..............................existsIndexs.add(0);
........................} else {
..............................existsIndexs.add(Integer.parseInt(substring.substring(1)));
........................}
..................}

..................if (null != treeNode.getChildren()) {
........................treeNode.getChildren().forEach(child - > self.go(child, self));
..................}

............};
............func.go(root, func);

............if (existsIndexs.isEmpty()) {
..................return srcName;
............}

............existsIndexs.sort(Comparator.comparingInt(o - > o));

............if (existsIndexs.get(0) != 0) {
..................return srcName;
............}

............for (int i = 1; i < existsIndexs.size(); i++) {
..................if (existsIndexs.get(i) - existsIndexs.get(i - 1) > 1) {
........................return String.format("%s_%d", srcName, i + 1);
..................}
............}

............return String.format("%s_%d", srcName, existsIndexs.size());
......}

}