关于java基础:JAVA基础之Collection的Lambda

明天讲一下Collection，以及Collection和Lamdba的联合

Collection包含
List（ArrayList，LinkedList）
Set（HashSet） — SortedSet（TreeSet）
Queue （PriorityQueue）—–Deque（LinkedList，ArrayDeque）
Map （HashMap）— SortedMap（TreeMap）（Map不属于Collection）
对于Iterable

boolean forEach(Cosumer<? super E> consumer);

对于Collection

boolean removeIf(Predicate<? super E> filter);

对于List

boolean replaceAll(UnaryOperator<? super E> operator);
boolean sort(Comparator<? super E> comparator);

public class Main {
    public static void main(String[] args) {
        List<BankAccount> bankAccountList = new ArrayList<>();
        bankAccountList.forEach(System.out::println);
        bankAccountList.removeIf(bankAccount -> bankAccount.getBalance()>20);
        List<String> names = new ArrayList<>();
        names.replaceAll(name->name.toUpperCase());
        names.replaceAll(String::toUpperCase);
        bankAccountList.sort(Comparator.comparing(BankAccount::getBalance)
                .thenComparing(BankAccount::getId));
    }
}

对于Map

void forEach(BiConsumer<? super K, ? super V> consumer);

Map<String,List<BankAccount>> map = new HashMap<>();
map.put("test",bankAccountList);
map.forEach((city,list)-> System.out.println(city+": "+list.size() +" account"));
输入：
test: 0 account

putIfAbsent办法，能够间接get后再间接调用add办法

Map<String, List<Person>> map = new HashMap<>();
map.putIfAbsent("boston",new ArrayList<>());
map.get("boston").add(new Person());

compute办法，返回的值为新值，同put办法相同，put办法返回的是旧值
computeIfAbsent办法，只有当key对应的value为空时才放入新值，并返回新值
computeIfPresent办法，只有当key对应的value为非空时才放入新值，并返回新值

V compute(K key, BiFunction<? super K, ? super V, ? extends V> remappingFunction)
V computeIfAbsent(K key, Function<? super K, ? extends V> mappingFunction) 
V computeIfPresent(K key, BiFunction<? super K, ? super V, ? extends V> remappingFunction) 
Map<String,String> map1 = new HashMap<>();
String val= map1.put("test","3");
System.out.println(val);
val= map1.compute("test", (k,v) -> "v");
System.out.println(val);
val = map1.computeIfAbsent("test",k -> "s");
System.out.println(val);
val = map1.computeIfPresent("test",(k,v)  -> "s1");
System.out.println(val);
输入
null
v
v
s1

computeIfAbsent办法，能够间接获取到新值并间接设置，写法简便

Map<String, Map<String,Person>> map = new HashMap<>();
map.computeIfAbsent("one",key -> new HashMap<String,Person>)
.put("two",john);

Map<String, List<Person>> map = new HashMap<>();
map.computeIfAbsent("one",key -> new ArrayList<Person>())
.add(john);

merge办法，获取原有key对应的value，通过Function操作，设置该key的值，并返回新值

V merge(K key, V value,
        BiFunction<? super V, ? super V, ? extends V> remappingFunction)
//接下面的将test的value设置为s1        
val =map1.merge("test","dd",(k,v)-> k+v);
System.out.println(val);
输入
s1dd

或者循环两个map的值，将其内容合并

Map<String,List<BankAccount>> map2 = new HashMap<>();
Map<String,List<BankAccount>> map3 = new HashMap<>();
map3.forEach(
        (key,value) ->
                map2.merge(key, value, (existingBankList, newBankList)->{
                    existingBankList.addAll(newBankList);
                    return existingBankList;
                })
);