如果我有值“foo”,和一个HashMap<String> ftw,其中ftw. containsvalue(“foo”)返回真,我怎么能得到相应的键?我需要遍历hashmap吗?最好的方法是什么?
当前回答
让我们看看我的例子
Map<String, String> mapPeopleAndCountry = new HashMap<>();
mapPeopleAndCountry.put("Matis", "Lithuania");
mapPeopleAndCountry.put("Carlos", "Honduras");
mapPeopleAndCountry.put("Teboho", "Lesotho");
mapPeopleAndCountry.put("Marielos", "Honduras");
List<String> peopleInHonduras = mapPeopleAndCountry.keySet()
.stream()
.filter(r -> mapPeopleAndCountry.get(r)
.equals("Honduras"))
.stream(Collectors.toList());
// will return ["Carlos", "Marielos"]
注:未经测试,可能含有错别字
其他回答
public static class SmartHashMap <T1 extends Object, T2 extends Object> {
public HashMap<T1, T2> keyValue;
public HashMap<T2, T1> valueKey;
public SmartHashMap(){
this.keyValue = new HashMap<T1, T2>();
this.valueKey = new HashMap<T2, T1>();
}
public void add(T1 key, T2 value){
this.keyValue.put(key, value);
this.valueKey.put(value, key);
}
public T2 getValue(T1 key){
return this.keyValue.get(key);
}
public T1 getKey(T2 value){
return this.valueKey.get(value);
}
}
使用薄包装:HMap
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
public class HMap<K, V> {
private final Map<K, Map<K, V>> map;
public HMap() {
map = new HashMap<K, Map<K, V>>();
}
public HMap(final int initialCapacity) {
map = new HashMap<K, Map<K, V>>(initialCapacity);
}
public boolean containsKey(final Object key) {
return map.containsKey(key);
}
public V get(final Object key) {
final Map<K, V> entry = map.get(key);
if (entry != null)
return entry.values().iterator().next();
return null;
}
public K getKey(final Object key) {
final Map<K, V> entry = map.get(key);
if (entry != null)
return entry.keySet().iterator().next();
return null;
}
public V put(final K key, final V value) {
final Map<K, V> entry = map
.put(key, Collections.singletonMap(key, value));
if (entry != null)
return entry.values().iterator().next();
return null;
}
}
我的2分钱。 您可以获取数组中的键,然后循环遍历数组。如果映射非常大,这将影响代码块的性能,因为首先获取数组中的键可能会消耗一些时间,然后再进行循环。否则,对于较小的地图,这应该是可以的。
String[] keys = yourMap.keySet().toArray(new String[0]);
for(int i = 0 ; i < keys.length ; i++){
//This is your key
String key = keys[i];
//This is your value
yourMap.get(key)
}
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
public class M{
public static void main(String[] args) {
HashMap<String, List<String>> resultHashMap = new HashMap<String, List<String>>();
Set<String> newKeyList = resultHashMap.keySet();
for (Iterator<String> iterator = originalHashMap.keySet().iterator(); iterator.hasNext();) {
String hashKey = (String) iterator.next();
if (!newKeyList.contains(originalHashMap.get(hashKey))) {
List<String> loArrayList = new ArrayList<String>();
loArrayList.add(hashKey);
resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
} else {
List<String> loArrayList = resultHashMap.get(originalHashMap
.get(hashKey));
loArrayList.add(hashKey);
resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
}
}
System.out.println("Original HashMap : " + originalHashMap);
System.out.println("Result HashMap : " + resultHashMap);
}
}
试试这个:
static String getKeyFromValue(LinkedHashMap<String, String> map,String value) {
for (int x=0;x<map.size();x++){
if( String.valueOf( (new ArrayList<String>(map.values())).get(x) ).equals(value))
return String.valueOf((new ArrayList<String>(map.keySet())).get(x));
}
return null;
}
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