/**
*(1) 找出2011年发生的所有交易,并按交易额排序(从低到高)。输出结果
- (2) 交易员都在哪些不同的城市工作过?
- (3) 查找所有来自于剑桥的交易员,并按姓名排序。
- (4) 返回所有交易员的姓名字符串,按字母顺序排序。
- (5) 有没有交易员是在米兰工作的?
- (6) 打印生活在剑桥的交易员的所有交易额。
- (7) 所有交易中,最高的交易额是多少?
- (8) 找到交易额最小的交易。
- 能使用方法引用的尽量用方法引用。
- /
使用Stream流完成上面问题
//交易员类
class Trader {
private String name;
private String city;
public Trader(String name, String city) {
this.name = name;
this.city = city;
}
public String getName() {
return name;
}
public String getCity() {
return city;
}
@Override
public String toString() {
return "Trader{" +
"name='" + name + '\'' +
", city='" + city + '\'' +
'}';
}
}
//Transaction(交易记录)
class Transaction {
private Trader trader; //交易员
private int year;
private int value;
public Transaction(Trader trader, int year, int value){
this.trader = trader;
this.year = year;
this.value = value;
}
public Trader getTrader(){
return this.trader;
}
public int getYear(){
return this.year;
}
public int getValue(){
return this.value;
}
public String toString(){
return "{" + this.trader + ", " +
"year: "+this.year+", " +
"value:" + this.value +"}";
}
};
public static void main(String[] args) {
Trader raoul = new Trader("Raoul", "Cambridge");
Trader mario = new Trader("Mario", "Milan");
Trader alan = new Trader("Alan", "Cambridge");
Trader brian = new Trader("Brian", "Cambridge");
List<Transaction> transactions = Arrays.asList(
new Transaction(brian, 2011, 300),
new Transaction(raoul, 2012, 1000),
new Transaction(raoul, 2011, 400),
new Transaction(mario, 2012, 710),
new Transaction(mario, 2012, 700),
new Transaction(alan, 2012, 950)
);