从给定的 JSON 返回给定级别的最高级别嵌套和项目列表

Question

如何从给定的 JSON 返回给定级别的最大嵌套级别和项目列表。我被要求不要使用任何第三方库。所以我尝试使用模式匹配，但无法得到结果。

{
"1" : {
"A" :{}
},
"2" : {
"A" :{}
"B" : {
"I" :{},
"II":{}
}
},
"3" : {}
}

我尝试过的代码：

public class NestedJson {
public int depth(String data, int deptth){
Pattern pattern = Pattern.compile(""(.+)"");
Matcher matcher = pattern.matcher(data);
List<String> list = new ArrayList<String>();
while (matcher.find()) {
list.add(matcher.group(1));
}
System.out.println(list.size());
return list.size();
}
public static void main(String args[]){
String data = "{n" +
"t"1" : {n" +
"t   "A" :{}n" +
"t},n" +
"t"2" : {n" +
"t   "A" :{}n" +
"t   "B" : {n" +
"t     "I" :{},n" +
"t     "II":{}n" +
"t   }n" +
"t},n" +
"t"3" : {}n" +
"}";
NestedJson nestedJson = new NestedJson();
nestedJson.depth(data, 1);
}
}

需要示例输出：

depth(data,1) -> {"max_level"= 3, items= [1,2,3]}
depth(data,2) -> {"max_level"= 3, items= [A,A,B]}

Answer 1

我会这样做，所以你不必再次扫描文本来要求不同的级别。

public final class NestedJson {
private int maxLevel;
private Map<Integer, List<String>> items = new HashMap<>();
public NestedJson(String data) {
int level = 0;
for (Matcher m = Pattern.compile(""([^"]*+)"|([{}])").matcher(data); m.find(); ) {
if (m.start(1) != -1) {
this.items.computeIfAbsent(level, k -> new ArrayList<>()).add(m.group(1));
if (level > this.maxLevel)
this.maxLevel = level;
} else if (m.group(2).charAt(0) == '{') {
level++;
} else {
level--;
}
}
}
public int getMaxLevel() {
return this.maxLevel;
}
public List<String> getItems(int level) {
return this.items.get(level);
}
}

测试

String data = "{n" +
"t"1" : {n" +
"t   "A" :{}n" +
"t},n" +
"t"2" : {n" +
"t   "A" :{}n" +
"t   "B" : {n" +
"t     "I" :{},n" +
"t     "II":{}n" +
"t   }n" +
"t},n" +
"t"3" : {}n" +
"}";
NestedJson nestedJson = new NestedJson(data);
System.out.println("maxLevel = " + nestedJson.getMaxLevel());
for (int level = 1; level <= nestedJson.getMaxLevel(); level++)
System.out.println("items(" + level + ") = " + nestedJson.getItems(level));

输出

maxLevel = 3
items(1) = [1, 2, 3]
items(2) = [A, A, B]
items(3) = [I, II]