使用复合键将树状地图转换为普通地图

Question

我试图在 Java 中做一些事情，但我被困在这个问题上，我真的不知道如何转换它。我尝试了 for 循环，但我需要做很多 for 循环，我想保持我的代码干净。

我有一个看起来像这样的地图。

Map 1:
  A: Object
  B: Object
  C: Map 2
     D: Object
     E: Object 
     F: Map 3
        G: Object
        (Here could be another map that could go really far, etc)
     H: Map 4
        I: Object
        J: Object
        (Here could also be more)\
     K: Object
     (Here could also be more)

请注意，这只是一个示例，映射、键等的数量可能不同，键名也可能不同。键名不能包含点。

C、F、H为地图，其中地图C包含D、E、F、H、K，地图F包含G等

如何将其转换为如下所示的一张地图：

A: Object
B: Object
C.D: Object
C.E: Object
C.K: Object
C.F.G: Object
C.H.I: Object
C.H.J: Object

键的顺序无关紧要。

Answer 1

像这样：

public Map<String, Object> convert (Map<String, Object> input, String key) {
    Map<String, Object> res = new HashMap<>();

    for (Entry<String, Object> e : input.entrySet()) {
        String newKey = key == "" ? e.getKey() : (key + "." + e.getKey());

        if (e.getValue() instanceof Map) {
            res.putAll(convert((Map) e.getValue(), newKey));  // recursive call
        } else {
            res.put(newKey, e.getValue());
        }
    }
    return res;
}

现在你在你的输入地图上调用它：

Map<String, Object> converted = convert(input, "");

Answer 2

沿着这些方向尝试一些东西假设您的地图是地图

    public class ObjectMapper
    {
    Map<Integer,Object> subMap;
    }

承认您的对象可以是字符串或映射假设您将 "path" 调用到这样的对象 “1#2#4#5”

    Map<Integer, Object> someMap;

    //assign map or load it from somewhere
    String searchPath = "1#2#4#5";
    String[] splitPath = searchPath .split("#");
    String searchResult = null;
    boolean resultFound = false;
    boolean resultWrongPlace = false;

    Map<Integer, Object> mapToSearch = null;
    for(int i = 0; i < splitPath.length; i++)
    {
    if (resultFound == true)
      {
         resultWrongPlace = true;
         break;
      }
      if (i == 0)
      {
        mapToSearch = someMap;
      }
      Object currentObj = mapToSearch.get(splitPath[Integer.parseInt(splitPath[i])])
      if (currentObj instanceof String)
      {
         searchResult = (String) currentObj ;
         resultFound = true;
      }
      else if (currentObj instanceof ObjectMapper)
      {
         mapToSearch = (ObjectMapper) currentObj ;
         resultFound = false;
      }
    }
    if (resultFound == true && resultWrongPlace == true)
    {
    throw new Exception("result was found but the path that lead to a  result was different from the indended path!");
    }
    else if (resultFound == false)
    {
    throw new Exception("no result was found searching your expression path!");
    }

我想说的是按照这些思路做的事情

使用复合键将树状地图转换为普通地图

Converting tree-like map into plain map with composite keys

java

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logic

dictionary