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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?

  •  1
     
  •   sxxkgoogle · 2020-04-10 22:04:59 +08:00 · 3325 次点击
    这是一个创建于 1674 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'
                    }
                ]
            }
        ]
    };
    

    请实现⼀一个查询函数,通过输⼊入 Tree 的 Root Node 和 Id,返回与其匹配的节点,函数原型如 下(注意:请不不要在函数内部直接⽤用 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:学⽣生按成绩分组 实现⼀一个 groupBy 函数,将学⽣生按照成绩等级进⾏行行分组。

    // 成绩等级分为 A 、B 和 C 三级
    function getGrade(score){
    return score < 60 ? 'C' : score < 80 ? 'B' : 'A';
    };
    // 学⽣生及其成绩
    let students = [
    {name: '张三', score: 84},
    {name: '李李四', 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:字符串串 Parse 请设计⼀一个字符串串 parse 函数,可以将输⼊入的字符串串分解为对应的树状结构,⽐比如:

    // 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'
    // }]
    // }]
    // };
    

    同理理,该 parse 函数可以分解如下任意情况,⽐比如:

    "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("typeArgs", substringJsonArr);
    
    	}
    
    }
    

    第四题:⾃自增 ID 已知有如下⼀一个树状数据结构:

    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,名字出现了了空隙,⽐比如 Tree 中有 button_1 和 button_3,但是没有 button_2,下⼀一 次插⼊入时,需要优先使⽤用 button_2,⽽而不不是 button_4 。 请实现⼀一个唯⼀一名称算法,当向整棵树的任意 children 插⼊入⼀一个新的 Node 时,可以保证 name 不不重复。 srcName 是计划向 Tree 中插⼊入节点的 name,⽐比如 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
        1
    mumbler  
       2020-04-10 22:09:38 +08:00 via Android
    代码都没有注释啊
    sxxkgoogle
        2
    sxxkgoogle  
    OP
       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。
    also24
        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
        4
    sxxkgoogle  
    OP
       2020-04-10 22:39:26 +08:00
    @also24 是全栈的岗位。
    mcfog
        5
    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
        6
    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
        7
    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
        8
    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
        9
    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
        10
    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
        11
    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
        12
    rabbbit  
       2020-04-11 03:36:06 +08:00   ❤️ 1
    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
        13
    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
        14
    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
        15
    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):
    ㅤㅤd = root and [root]
    ㅤㅤwhile d:
    ㅤㅤㅤㅤr = next((r for r in d if r['id'] == id), None)
    ㅤㅤㅤㅤif r:
    ㅤㅤㅤㅤㅤㅤreturn r
    ㅤㅤㅤㅤd = sum((r['children'] for r in d if 'children' in r), [])
    ㅤㅤreturn None

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

    语法题栈实现
    def parse(args):
    ㅤㅤstack = [{'type': ''}]
    ㅤㅤfor c in args:
    ㅤㅤㅤㅤif c == ' ':
    ㅤㅤㅤㅤㅤㅤcontinue
    ㅤㅤㅤㅤelif c == '<':
    ㅤㅤㅤㅤㅤㅤstack[-1]['typeArgs'] = [{'type': ''}]
    ㅤㅤㅤㅤㅤㅤstack.append(stack[-1]['typeArgs'][0])
    ㅤㅤㅤㅤelif c == ',':
    ㅤㅤㅤㅤㅤㅤstack[-2]['typeArgs'].append({'type': ''})
    ㅤㅤㅤㅤㅤㅤstack[-1] = stack[-2]['typeArgs'][-1]
    ㅤㅤㅤㅤelif c == '>':
    ㅤㅤㅤㅤㅤㅤdel stack[-1]
    ㅤㅤㅤㅤelse:
    ㅤㅤㅤㅤㅤㅤstack[-1]['type'] += c
    ㅤㅤreturn 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):
    ㅤㅤd = set()
    ㅤㅤdef dfs(r):
    ㅤㅤㅤㅤr['name'].rsplit('_', 1)[0] == srcName and d.add(r['name'])
    ㅤㅤㅤㅤif 'children' in r:
    ㅤㅤㅤㅤㅤㅤfor child in r['children']:
    ㅤㅤㅤㅤㅤㅤㅤㅤdfs(child)
    ㅤㅤrootTreeNode and dfs(rootTreeNode)
    ㅤㅤif srcName not in d:
    ㅤㅤㅤㅤreturn srcName
    ㅤㅤfor i in range(1, len(d) + 1):
    ㅤㅤㅤㅤres = srcName + '_' + str(i)
    ㅤㅤㅤㅤif res not in d:
    ㅤㅤㅤㅤㅤㅤreturn res
    rabbbit
        16
    rabbbit  
       2020-04-11 05:24:58 +08:00
    @lithbitren 空格是怎么打出来的?
    lithbitren
        17
    lithbitren  
       2020-04-11 05:34:27 +08:00   ❤️ 1
    @rabbbit 随便找了个空白字符,ascii 码 12644
    tyx1703
        18
    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
        19
    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
        20
    yazoox  
       2020-04-11 08:10:40 +08:00 via Android
    有意思,关注一下
    alphatoad
        21
    alphatoad  
       2020-04-11 08:11:25 +08:00
    别想太多,可能只是因为面试有 kpi
    easylee
        22
    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
        23
    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
        24
    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
        25
    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 {

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    }
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