我需要按對象的 id 合并數(shù)組。我的班級有兩個數(shù)組（本地和全局）：public class Info { private int id;// never change private String msg; private boolean isFavor;// presents only in local array// constructor,getters, setters etc }合并規(guī)則：如果對象不存在于全局數(shù)組中，我們不合并它；僅從全局對象中獲取 msg ；如果對象存在于全局數(shù)組中，我們添加（合并）它；變量“isFavor”取自本地對象；數(shù)組沒有排序；例子：local arayList = {Info(1,"msg1",false),Info(2,"msgTwo",false),Info(3,"msg3",true), Info(4,"msg4",true)};global arayList = {Info(1,"msg1",false),Info(2,"msg2",false),Info(3,"msg3",false),Info(5,"msg5",false)}result arayList = {Info(1,"msg1",false),Info(2,"msg2",false),Info(3,"msg3",true),Info(5,"msg5",false)}
2 回答
慕勒3428872
TA貢獻1848條經(jīng)驗 獲得超6個贊
所以，我做了這樣的事情，我認為它可以滿足你的要求?；旧希宜龅氖且韵聝?nèi)容。我將所有全局值添加到映射中，并在局部值的下一次迭代中檢查 id 是否存在，然后更改isFavor.
public static void main(String[] args) throws ParseException {
List<Info> local = new ArrayList<>();
local.add(new Info(1,"msg1",false));
local.add(new Info(2,"msgTwo",false));
local.add(new Info(3,"msg3",true));
local.add(new Info(4,"msg4",true));
List<Info> global = new ArrayList<>();
global.add(new Info(1,"msg1",false));
global.add(new Info(2,"msg2",false));
global.add(new Info(3,"msg3",false));
global.add(new Info(4,"msg5",true));
Map<Integer, Info> map = new HashMap<>();
for (Info info : global) {
if (!map.containsKey(info.id)) {
map.put(info.id, info);
}
}
for (Info info : local) {
if (map.containsKey(info.id)){
map.get(info.id).isFavor = info.isFavor;
}
}
System.out.println(map.values());
}
鴻蒙傳說
TA貢獻1865條經(jīng)驗 獲得超7個贊
public Info [] merge(Info [] localArray, Info [] globalArray) {
List<Info> resultList = new ArrayList();
for(Info infoGlobal : globalArray) {
Info infoLocal = findInfo(infoGlobal.getId(), localArray);
if( infoLocal != null )
infoGlobal.setFavor(infoLocal.isFavor());
resultList.add(infoGlobal);
}
return resultList.toArray();
}
private Info findInfo(int id, Info [] infoArray) {
for(Info info : infoArray)
if(info.getId == id)
return info ;
return null ;
}
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