我有課：class MyClass { String s1; String s2;}列表對(duì)象包括[{s1: "a", s2: "b"}, {s1: "a", s2: "c"}, {s1: "a", s2: "b"}, {s1: "a", s2：“d”}]我想刪除對(duì)象重復(fù)的 2 個(gè)屬性 s1 和 s2預(yù)期結(jié)果：[{s1: "a", s2: "b"}, {s1: "a", s2: "c"}, {s1: "a", s2: "d"}]如果能用Java 8更好幫我 ：（
3 回答
瀟湘沐
TA貢獻(xiàn)1816條經(jīng)驗(yàn) 獲得超6個(gè)贊
首先像這樣覆蓋equals和hashCode方法MyClass，
@Override
public int hashCode() {
return 31 * s1.hashCode() + s2.hashCode();
}
@Override
public boolean equals(Object obj) {
return obj instanceof MyClass && ((MyClass) obj).getS1().equals(getS1())
&& ((MyClass) obj).getS2().equals(getS2());
}
那么消除重復(fù)的最簡(jiǎn)單方法是使用Set.
new HashSet<>(instances)
慕標(biāo)琳琳
TA貢獻(xiàn)1830條經(jīng)驗(yàn) 獲得超9個(gè)贊
實(shí)現(xiàn) hashcode() 或 equals() 不是必須的。這是一種方法：
Map<List<String>, MyClass> map = new HashMap<>();
List<MyClass> distinctList = myCollection.stream()
.collect(Collectors.toMap(c -> Arrays.asList(c.s1, c.s2), Function.identity(), (x,y)->x))
.values()
.stream()
.collect(Collectors.toList());
手掌心
TA貢獻(xiàn)1942條經(jīng)驗(yàn) 獲得超3個(gè)贊
您需要覆蓋equal和hash code函數(shù)，然后使用java 8的Streaming API的distinct函數(shù)。
import java.util.ArrayList;
import java.util.List;
import java.util.Objects;
import java.util.stream.Collectors;
public class MyClass {
private String s1;
private String s2;
public MyClass() {
}
public MyClass(String s1, String s2) {
this.s1 = s1;
this.s2 = s2;
}
public String getS1() {
return s1;
}
public MyClass setS1(String s1) {
this.s1 = s1;
return this;
}
public String getS2() {
return s2;
}
public MyClass setS2(String s2) {
this.s2 = s2;
return this;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
MyClass myClass = (MyClass) o;
return Objects.equals(getS1(), myClass.getS1()) &&
Objects.equals(getS2(), myClass.getS2());
}
@Override
public int hashCode() {
return Objects.hash(getS1(), getS2());
}
@Override
public String toString() {
return "MyClass{" + "s1='" + s1 + '\'' + ", s2='" + s2 + '\'' + '}';
}
public static void main(String[] args) {
List<MyClass> list = new ArrayList<>();
list.add(new MyClass("a", "b"));
list.add(new MyClass("a", "c"));
list.add(new MyClass("a", "b"));
list.add(new MyClass("a", "d"));
System.out.println("Original size is : " + list.size()); // Size is 4
list.forEach(System.out::println);
//Now remove duplicate
List<MyClass> expected = list.stream().distinct().collect(Collectors.toList());
System.out.println(
"After remove duplication size is : " + expected.size()); // Here is your expected result size is 3
expected.forEach(System.out::println);
}
}
這是輸出：
原尺寸為：4
我的類{s1='a', s2='b'}
我的類{s1='a', s2='c'}
我的類{s1='a', s2='b'}
我的類{s1='a', s2='d'}
刪除重復(fù)后大小為：3
我的類{s1='a', s2='b'}
我的類{s1='a', s2='c'}
我的類{s1='a', s2='d'}
添加回答
舉報(bào)
0/150
提交
取消