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nullvvs1

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;                }));