我需要根據父子關系將數據列表轉換為列表列表。如果 parent 為 null 落入一級，則二級將基于一級 id。我的數據如下所示：[ {id:1, parent: null }, {id:2, parent: 1 }, {id:3, parent: 1 }, {id:4, parent: 1 }, {id:5, parent: 2 }, {id:6, parent: 2 }, {id:7, parent: 3 }, {id:8, parent: 3 }, {id:9, parent: 4 }, {id:10, parent: 4 }, {id:11, parent: 5 }, {id:12, parent: null }, {id:13, parent: 12 },]我的代碼是：響應數據Map<String,Map<String,ResponseData>> map = new HashMap<>();for (ResponseData responseData : responseDataList) { Map<String,responseData> responseDatasMap = map.get(responseData.getParent()); if(responseDatasMap != null) { responseDatasMap.put(responseData.getId(),responseData); map.put(responseData.getParent(),responseDatasMap); } else { responseDatasMap = new HashMap<>(); responseDatasMap.put(responseData.getParent(),responseData); map.put(responseData.getParent(),responseDatasMap); }}上面的地圖將包含父級作為鍵和映射到父級的值的地圖List<List<ResponseData>> sections = new ArrayList<>();for (Map.Entry<String,Map<String, ResponseData>> responseDataMap : map.entrySet()) { Map<String, ResponseData> valueMap = responseDataMap.getValue(); responseDataList = new ArrayList<>(); for(Map.Entry<String, ResponseData> responseData :valueMap.entrySet()) { responseDataList.add(responseData.getValue()); } sections.add(responseDataList);}我的輸出如下所示：[ [ {id:1, parent: null } ], [ {id:2, parent: 1 },{id:3, parent: 1 },{id:4, parent: 1 } ], [ {id:5, parent: 2 },{id:6, parent: 2 } ], [ {id:7, parent: 3 },{id:8, parent: 3 } ], [ {id:9, parent: 4 },{id:10, parent: 4 } ], [ {id:11, parent: 5 }]]請檢查并讓我知道我們如何實施。提前致謝
1 回答
PIPIONE
TA貢獻1829條經驗 獲得超9個贊
為了表示樹結構，我使用了ArrayList，其中節點的索引等于其在數組中的索引 + 1。如果您有稀疏樹/某些索引可能丟失，請使用映射的等效方法。
使用 Java 8 流 API 的解決方案：
public static void main( String[] args ) {
List<ResponseData> responseDataList = Arrays.asList(
new ResponseData( 1, -1 ), // changed null to -1 as null can't be a map key
new ResponseData( 2, 1 ),
new ResponseData( 3, 1 ),
new ResponseData( 4, 1 ),
new ResponseData( 5, 2 ),
new ResponseData( 6, 2 ),
new ResponseData( 7, 3 ),
new ResponseData( 8, 3 ),
new ResponseData( 9, 4 ),
new ResponseData( 10, 4 ),
new ResponseData( 11, 5 ),
new ResponseData( 12, -1 ),
new ResponseData( 13, 12 )
);
final Map<Integer, List<ResponseData>> map = responseDataList.stream()
.collect( Collectors.groupingBy( o -> getLevel( responseDataList, o, 0 ) ) );
System.out.println( map );
// To convert the Map to a List of Lists:
System.out.println( new ArrayList<>( map.values() ));
}
private static int getLevel(List<ResponseData> nodes, ResponseData responseData, int level) {
if( responseData.parent == -1 ) {
return level;
} else {
return getLevel( nodes, nodes.get( responseData.parent - 1 ), level + 1 ); // -1 to adjust index
}
}
private static final class ResponseData {
public int id;
public int parent;
public ResponseData( int id, int parent ) {
this.id = id;
this.parent = parent;
}
@Override
public String toString() {
return String.format( "{id: %d, parent: %d}", id, parent );
}
}
此外，此代碼期望您的樹確實是一棵樹。如果有任何循環，它將無限循環，最終因堆棧溢出而失敗。
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