Biegral Blog

1.获取部门列表

2.根据部门的Id映射部门实体类

先看下部门实体类

public class TbEnterpriseDepartment {
    private int id;
    private String name;
    private int number;
    private int parentId;
    private List<TbEnterpriseDepartment> childDeparents;
    public TbEnterpriseDepartment() {
    }

    public TbEnterpriseDepartment(int id, String name, int parentId, boolean isRoot) {
        this.id = id;
        this.name = name;
        this.parentId = parentId;
        this.isRoot = isRoot;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getNumber() {
        return number;
    }

    public void setNumber(int number) {
        this.number = number;
    }

    public int getParentId() {
        return parentId;
    }

    public void setParentId(int parentId) {
        this.parentId = parentId;
    }

    @Override
    public String toString() {
        return JSON.toJSONString(this);
    }

    public List<TbEnterpriseDepartment> getChildDeparents() {
        return childDeparents;
    }

    public void setChildDeparents(List<TbEnterpriseDepartment> childDeparents) {
        this.childDeparents = childDeparents;
    }
 //添加子部门
    public void addChildDepartment(TbEnterpriseDepartment department) {
        if(this.childDeparents == null)
            this.childDeparents = new LinkedList<>();
        this.childDeparents.add(department);
    }
}

3.然后遍历之前的部门列表,把部门加入到父部门的child列表中。

4.打印输出JSON

5.代码

@Test
    public void test() {
        //初始化列表
        List<TbEnterpriseDepartment> departments = new ArrayList<>();
        TbEnterpriseDepartment rootDepartment = new TbEnterpriseDepartment(1,"root-1",-1,true);
        departments.add(new TbEnterpriseDepartment(2,"root-1-1",1,false));
        departments.add(new TbEnterpriseDepartment(3,"root-1-1-1",2,false));
        departments.add(new TbEnterpriseDepartment(4,"root-1-1-2",2,false));
        departments.add(new TbEnterpriseDepartment(5,"root-1-2",1,false));
        departments.add(new TbEnterpriseDepartment(6,"root-1-2-1",5,false));
        departments.add(rootDepartment);

        HashMap<Integer,TbEnterpriseDepartment>departmentMap = new HashMap<>();

        //初始化HashMap
        for (TbEnterpriseDepartment department : departments) {
            departmentMap.put(department.getId(),department);
        }
        //遍历添加
        for (TbEnterpriseDepartment department : departments) {
//如果确定parentId一定存在去除此条件            if(departmentMap.containsKey(department.getParentId())){
                departmentMap.get(department.getParentId()).addChildDepartment(department);
            }
        }
        System.out.println(rootDepartment);
        //结果
//{"childDepartments":[{"childDepartments":[{"id":3,"name":"root-1-1-1","parentId":2},{"id":4,"name":"root-1-1-2","parentId":2}],"id":2,"name":"root-1-1","parentId":1},{"childDepartments":[{"id":6,"name":"root-1-2-1","parentId":5}],"id":5,"name":"root-1-2","parentId":1}],"id":1,"name":"root-1","parentId":-1}
    }    

在HashMap的理想情况下复杂度为O（n）