我有一个动态检索的层次结构树(通过REST服务)。我根据我想要的深度和数据的限制来限制数据。我想把那棵树夷为平地,所以首先是孩子,然后是孙子,等等。
例如:
1
-2
-4
-5
-8
-3
-6
-7
-9如果深度为100,极限为100,则应为2 3 4 5 6 7 8 9
如果深度为1,极限为100,则应为2 3。
如果深度为2,极限为5,则应为2 3 4 5 6
现在,,我有一个递归算法,但是它不是按级别平缓,而是递归地(2,4,5,8,3,7,9)。
以下是实际代码:
@GET
@Path("/GetDatas")
public Response getDatas(@QueryParam("clientId") final String clientId,
@QueryParam("maxDepth") final Integer maxDepth,
@QueryParam("limit") final Integer limit) {
Set datas = new LinkedHashSet();
findChildren(clientId, maxDepth, limit, datas, 0);
return Response.status(Response.Status.OK).entity(datas).build();
}
private void findChildren(String clientId, Integer maxDepth, Integer limit, Set datas, Integer actualDepth) {
// here we are getting the data via a REST WS
results = .... (function(clientId))
for (final String result : results) {
if (datas.size() < limit) {
if (!datas.contains(result)) {
datas.add(result);
if (actualDepth < maxDepth) {
findChildren(result, maxDepth, limit, datas, actualDepth + 1);
}
}
}
}
}我简化了一点。实际上,在现实中,节点也将自己作为孙辈( getChildren将根据算法检索匹配数据,因此如果2是1的匹配,则1是2的匹配)。
清单的顺序也很重要。
下面是JDoodle,以便您可以测试:jdoodle.com/ia/gFm
发布于 2021-09-09 07:53:19
下面的mre使用BFS将树夷为平地,同时遵守限制:
import java.util.*;
public class MyClass {
public static void main(String[] args) {
new DataClass().execute();
}
static class DataClass {
public void execute() {
Map<String, List<String>> tree = new LinkedHashMap();
tree.put("1", Arrays.asList("2", "3"));
tree.put("2", Arrays.asList("4", "5"));
tree.put("3", Arrays.asList("6", "7"));
tree.put("5", Arrays.asList("8"));
tree.put("7", Arrays.asList("9"));
tree.put("4", Arrays.asList());
tree.put("6", Arrays.asList());
tree.put("8", Arrays.asList());
tree.put("9", Arrays.asList());
int maxDepth =100, maxNodes =100;
System.out.println("max depth:"+ maxDepth + " max nodes:"+ maxNodes +" - "+ findChildren(maxDepth, maxNodes, tree));
maxDepth =1; maxNodes =100;
System.out.println("max depth:"+ maxDepth + " max nodes:"+ maxNodes +" - "+ findChildren(maxDepth, maxNodes, tree));
maxDepth =2; maxNodes =5;
System.out.println("max depth:"+ maxDepth + " max nodes:"+ maxNodes +" - "+findChildren(maxDepth, maxNodes, tree));
}
//helper method for bfs
Set<String> findChildren(int maxDepth, int maxNodes, Map<String, List<String>> tree) {
Set<String> flatTree = new LinkedHashSet<>(); //hold and return the flatten tree
final String root = "1";
List<String> nodesAtCurrentDepth = new ArrayList<>();//hold all nodes of the current depth
nodesAtCurrentDepth.add(root);
return findChildren(maxDepth, maxNodes, 0, flatTree, nodesAtCurrentDepth, tree);
}
//flatten tree using bfs
Set<String> findChildren(int maxDepth, int maxNodes, int currentDepth, Set<String> flatTree,
List<String> nodesAtCurrentDepth, Map<String, List<String>> tree) {
if(currentDepth < maxDepth && ! nodesAtCurrentDepth.isEmpty()) {
List<String> nodesAtNextDepth = new ArrayList<>();//collects all next level nodes
//add all next depth nodes to nodesAtNextDepth, respecting maxNodes limit
for(String node : nodesAtCurrentDepth){
for(String childNode : tree.get(node)){
if(flatTree.size() + nodesAtNextDepth.size() >= maxNodes) {
break;
}
nodesAtNextDepth.add(childNode);
}
}
flatTree.addAll(nodesAtNextDepth);
currentDepth++;
nodesAtCurrentDepth = new ArrayList<>(nodesAtNextDepth);
findChildren(maxDepth, maxNodes, currentDepth, flatTree, nodesAtCurrentDepth, tree);
};
return flatTree;
}
}
}输出:
最大深度:100个最大节点:100- 2,3,4,5,6,7,8,9 最大深度:1最大节点:100- 2,3 最大深度:2个最大节点:5- 2,3,4,5,6
https://stackoverflow.com/questions/69105361
复制相似问题