请教一下关于 lamdba 的用法

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@Data
class User {
    Integer id;
    Integer parentId;
    List<User> userList;

    public User(Integer id, Integer parentId) {
        this.id = id;
        this.parentId = parentId;
    }

    public static void main(String[] args) {
        List<User> userList = new ArrayList<>();
        userList.add(new User(1, null));
        userList.add(new User(2, 1));
        userList.add(new User(3, 1));
    }
}

请教一下怎样使用 lamdba 将 u2u3 set 到 u1 里
用 lamdba 总用不好

Supplement 1 · Mar 22, 2018

            List<User> rootList = userResource.stream().filter(u-> u.getParentId() == null).collect(toList());
            Set<User> userSet= new HashSet<>();
            for(User u: rootList){
               List<User> newList = new ArrayList<>();
                for(User subUser: userResource){
                    if(subUser.getParentId == u.getId){
                         newList.add(subUser);
                         u.setUserList(newList);
                    }
                }
                userSet.add(u);
            }

目前我用拆分成两个List rootList 代表根节点，userResource是所有数据

11 replies • 2018-05-24 17:54:31 +08:00

MinQ

Mar 22, 2018

等等不是 lambda 么？

yidinghe

Mar 22, 2018

这个用不到 lambda 表达式，但我猜你是想要这样：

```java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class User {

private int id;

private int parentId;

private List<User> userList = new ArrayList<>();

public User(int id, User... users) {
this.id = id;
this.userList.addAll(Arrays.asList(users));
this.userList.forEach(u -> u.parentId = id);
}

public static void main(String[] args) {
User root = new User(1,
new User(2,
new User(3),
new User(4)),
new User(5,
new User(6)
)
);
}
}
```

yidinghe

Mar 22, 2018

为什么回复不能用 markdown........
算了贴到别地方 https://gitee.com/yidinghe/codes/u3w7r5qz2ikxp9hotcd1n41

zhazi

Mar 22, 2018

@yidinghe 结构是这样的，但是我数据是从数据库取的一个 List<User>里面有 parent 节点和 child 节点，我不想一层又一层遍历，我觉得 stream 应该能处理这样的需求

yidinghe

Mar 22, 2018

@zhazi 没看懂你到底想得到什么，要不你先用标准写法写完，然后再看看怎么改。

zhazi

Mar 22, 2018

@yidinghe 我在主题上补充了一下，麻烦你看下，谢谢

yidinghe

Mar 22, 2018

哦这就是将集合元素构建成树结构，我这里给个现成的例子：
https://gitee.com/yidinghe/codes/q6lbevnaru5m08fixcw4740

leafin

Mar 22, 2018

试试这个
List<User> userList = userResource.stream()
.peek(u -> u.setUserList(userResource.stream()
.filter(su -> Objects.equals(u.getId(), su.getParentId()))
.collect(toList())))
.filter(u -> u.getParentId() == null)
.collect(toList());

zacard

Mar 22, 2018

private Set<User> something(List<User> userResource) {
Map<Integer, List<User>> groupMap = userResource.stream()
.collect(Collectors.groupingBy(User::getParentId, Collectors.toList()));
Map<Integer, User> rootMap = groupMap.remove(null).stream()
.collect(Collectors.toMap(User::getId, Function.identity()));
groupMap.forEach((k, v) -> rootMap.get(k).setUserList(v));
return new HashSet<>(rootMap.values());
}

没有测试过，不过大致是这个思路

waltyyy

Mar 29, 2018

```
Map<Integer, List<User>> parentIdUserMap = userResource.stream()
.filter(u -> u.getParentId() != null)
.collect(groupingBy(User::getParentId));

userResource.stream().filter(u -> u.getParentId() == null).map(u -> {
u.setUserList(parentIdUserMap.get(u.getId()));
return u;
}).collect(toList());
```

ng14

May 24, 2018 via Android

@yidinghe 我觉得差不多