我需要根據(jù)父子關(guān)系將數(shù)據(jù)列表轉(zhuǎn)換為列表列表。如果 parent 為 null 落入一級(jí)，則二級(jí)將基于一級(jí) id。我的數(shù)據(jù)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(yīng)數(shù)據(jù)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(jí)作為鍵和映射到父級(jí)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(qǐng)檢查并讓我知道我們?nèi)绾螌?shí)施。提前致謝