Java 层级显示的集合?
Java Collection for hierarchy display?
我正在尝试找到一种方法来使用 table 中的当前数据结构创建资产管理报告:
Company_Name Company_id Managing_Company_id
Target 002 150
Coles 003 150
Wesfarmers 150 100
Shell 001 150
Max Company X 100 null
Woolworths Group 250 100
Caltex 201 250
Woolworths 202 250
Aldi 300 null
PGX 650 600
并且报告会将其迭代到层次结构中
Max Company X
Wesfarmers
Target
Coles
Shell
Woolworths Group
Caltex
Woolworths
Aldi
PGX
是否有允许我管理此结构的集合?
关于使用正确数据结构的问题是主观的。即使以目前的数据结构,获取你需要的信息应该不难。您也可以选择 List
的 Tree
。另一方面,您可以轻松地旋转自己的数据结构,例如。自定义 tree
并将每个根节点存储在 HashMap
中。也就是说,Java 不提供存储此类特殊场景的直接方式。如果你有一个大数据集,你应该考虑使用某种基于图形的数据库。
乍一看,我正在考虑 topological sort
用 graph/tree 来处理你的情况,但当我正在实施解决方案时。我认为仅使用 double 映射记录关系没有必要解决您的问题:
id
和name
的关系;
managing
和managed
的关系;
然后我们可以有一个解决方案:
public static void main(String args[]) {
List<List<String>> listList = prepareTestData();
Map<String, String> idNameMap = new HashMap<>();
Map<String, String> managingRelationMap = new HashMap<>();
List<String> theTop = new ArrayList<>();
for (List<String> list : listList) {
idNameMap.put(list.get(1), list.get(0));
if (list.get(2) == null || list.get(2).equals(list.get(1))) { // no managing or managed by itself;
theTop.add(list.get(1));
} else {
managingRelationMap.put(list.get(1), list.get(2));
}
}
// add specials whose managing does not exist;
for (String s : managingRelationMap.keySet()) {
if (!idNameMap.containsKey(managingRelationMap.get(s))) {
theTop.add(s);
}
}
for (String top : theTop) {
printManagingStructure(top, managingRelationMap, idNameMap, 1);
}
}
private static List<List<String>> prepareTestData() {
List<List<String>> listList = new ArrayList<>();
listList.add(new ArrayList<>(Arrays.asList("Target", "002", "150")));
listList.add(new ArrayList<>(Arrays.asList("Coles", "003", "150")));
listList.add(new ArrayList<>(Arrays.asList("Wesfarmers", "150", "100")));
listList.add(new ArrayList<>(Arrays.asList("Shell", "001", "150")));
listList.add(new ArrayList<>(Arrays.asList("Max Company X", "100", null)));
listList.add(new ArrayList<>(Arrays.asList("Woolworths Group", "250", "100")));
listList.add(new ArrayList<>(Arrays.asList("Caltex", "201", "250")));
listList.add(new ArrayList<>(Arrays.asList("Woolworths", "202", "250")));
listList.add(new ArrayList<>(Arrays.asList("Aldi", "300", null)));
listList.add(new ArrayList<>(Arrays.asList("PGX", "650", "600")));
return listList;
}
private static void printManagingStructure(String managingId, Map<String, String> managingRelationMap,
Map<String, String> idNameMap, int gapCount) {
System.out.println(String.format("%" + gapCount * 20 + "s%-20s", "", idNameMap.get(managingId)));
if (!managingRelationMap.containsValue(managingId)) {
return;
}
managingRelationMap.entrySet().stream().filter(entry -> entry.getValue().equals(managingId))
.forEach(entry -> {
printManagingStructure(entry.getKey(), managingRelationMap, idNameMap, gapCount + 1);
});
}
输出将是:
Max Company X
Wesfarmers
Shell
Target
Coles
Woolworths Group
Caltex
Woolworths
Aldi
PGX
我正在尝试找到一种方法来使用 table 中的当前数据结构创建资产管理报告:
Company_Name Company_id Managing_Company_id
Target 002 150
Coles 003 150
Wesfarmers 150 100
Shell 001 150
Max Company X 100 null
Woolworths Group 250 100
Caltex 201 250
Woolworths 202 250
Aldi 300 null
PGX 650 600
并且报告会将其迭代到层次结构中
Max Company X
Wesfarmers
Target
Coles
Shell
Woolworths Group
Caltex
Woolworths
Aldi
PGX
是否有允许我管理此结构的集合?
关于使用正确数据结构的问题是主观的。即使以目前的数据结构,获取你需要的信息应该不难。您也可以选择 List
的 Tree
。另一方面,您可以轻松地旋转自己的数据结构,例如。自定义 tree
并将每个根节点存储在 HashMap
中。也就是说,Java 不提供存储此类特殊场景的直接方式。如果你有一个大数据集,你应该考虑使用某种基于图形的数据库。
乍一看,我正在考虑 topological sort
用 graph/tree 来处理你的情况,但当我正在实施解决方案时。我认为仅使用 double 映射记录关系没有必要解决您的问题:
id
和name
的关系;managing
和managed
的关系;
然后我们可以有一个解决方案:
public static void main(String args[]) {
List<List<String>> listList = prepareTestData();
Map<String, String> idNameMap = new HashMap<>();
Map<String, String> managingRelationMap = new HashMap<>();
List<String> theTop = new ArrayList<>();
for (List<String> list : listList) {
idNameMap.put(list.get(1), list.get(0));
if (list.get(2) == null || list.get(2).equals(list.get(1))) { // no managing or managed by itself;
theTop.add(list.get(1));
} else {
managingRelationMap.put(list.get(1), list.get(2));
}
}
// add specials whose managing does not exist;
for (String s : managingRelationMap.keySet()) {
if (!idNameMap.containsKey(managingRelationMap.get(s))) {
theTop.add(s);
}
}
for (String top : theTop) {
printManagingStructure(top, managingRelationMap, idNameMap, 1);
}
}
private static List<List<String>> prepareTestData() {
List<List<String>> listList = new ArrayList<>();
listList.add(new ArrayList<>(Arrays.asList("Target", "002", "150")));
listList.add(new ArrayList<>(Arrays.asList("Coles", "003", "150")));
listList.add(new ArrayList<>(Arrays.asList("Wesfarmers", "150", "100")));
listList.add(new ArrayList<>(Arrays.asList("Shell", "001", "150")));
listList.add(new ArrayList<>(Arrays.asList("Max Company X", "100", null)));
listList.add(new ArrayList<>(Arrays.asList("Woolworths Group", "250", "100")));
listList.add(new ArrayList<>(Arrays.asList("Caltex", "201", "250")));
listList.add(new ArrayList<>(Arrays.asList("Woolworths", "202", "250")));
listList.add(new ArrayList<>(Arrays.asList("Aldi", "300", null)));
listList.add(new ArrayList<>(Arrays.asList("PGX", "650", "600")));
return listList;
}
private static void printManagingStructure(String managingId, Map<String, String> managingRelationMap,
Map<String, String> idNameMap, int gapCount) {
System.out.println(String.format("%" + gapCount * 20 + "s%-20s", "", idNameMap.get(managingId)));
if (!managingRelationMap.containsValue(managingId)) {
return;
}
managingRelationMap.entrySet().stream().filter(entry -> entry.getValue().equals(managingId))
.forEach(entry -> {
printManagingStructure(entry.getKey(), managingRelationMap, idNameMap, gapCount + 1);
});
}
输出将是:
Max Company X
Wesfarmers
Shell
Target
Coles
Woolworths Group
Caltex
Woolworths
Aldi
PGX