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此代码说明了您想要的一切。注意 hashCode、equals 和 compareTo 方法是如何被覆盖的。package com.pkr.test;import java.text.SimpleDateFormat;import java.util.ArrayList;import java.util.Collections;import java.util.Date;import java.util.HashSet;import java.util.List;import java.util.Set;public class User implements Comparable<User> { public static SimpleDateFormat sdf = new SimpleDateFormat("dd-MM-yyyy"); private long userId; private String userName; private Date createdDate; public User(long userId, String userName, Date createdDate) { this.userId = userId; this.userName = userName; this.createdDate = createdDate; } public long getUserId() { return userId; } public void setUserId(long userId) { this.userId = userId; } public String getUserName() { return userName; } public void setUserName(String userName) { this.userName = userName; } public Date getCreatedDate() { return createdDate; } public void setCreatedDate(Date createdDate) { this.createdDate = createdDate; } public String toString() { return String.format("userId: %s, userName: %s, createdDate: %s", new Object[] { userId, userName, sdf.format(createdDate) }); } public boolean equals(Object o) { if (o instanceof User) { return ((User) o).getUserId() == this.getUserId(); } return false; } public int hashCode() { return new Long(userId).hashCode(); } @Override public int compareTo(User o) { return this.getCreatedDate().compareTo(o.getCreatedDate()); } public static void main(String args[]) throws Exception { Set<User> userSet = new HashSet<User>(); userSet.add(new User(1, "Pushpesh", sdf.parse("16-02-2018"))); userSet.add(new User(2, "Vikrant", sdf.parse("12-02-2018"))); userSet.add(new User(3, "Abhay", sdf.parse("11-02-2018"))); userSet.add(new User(4, "Komal", sdf.parse("18-02-2018"))); userSet.stream().forEach(System.out::println); // this will remove user with id 2 no matter what his name and // createdDate are userSet.remove(new User(2, "Vikrant", sdf.parse("12-02-2018"))); System.out.println("After removing userId 2"); userSet.stream().forEach(System.out::println); System.out.println(); List<User> userList = new ArrayList<User>(userSet); System.out.println("Before sorting"); userList.stream().forEach(System.out::println); Collections.sort(userList); // This will sort based on date System.out.println("After sorting"); userList.stream().forEach(System.out::println); }}希望这能澄清你想要的一切。如果您有任何疑问，请告诉我。执行后，此代码产生以下输出，userId: 1, userName: Pushpesh, createdDate: 16-02-2018userId: 2, userName: Vikrant, createdDate: 12-02-2018userId: 3, userName: Abhay, createdDate: 11-02-2018userId: 4, userName: Komal, createdDate: 18-02-2018After removing userId 2userId: 1, userName: Pushpesh, createdDate: 16-02-2018userId: 3, userName: Abhay, createdDate: 11-02-2018userId: 4, userName: Komal, createdDate: 18-02-2018Before sortinguserId: 1, userName: Pushpesh, createdDate: 16-02-2018userId: 3, userName: Abhay, createdDate: 11-02-2018userId: 4, userName: Komal, createdDate: 18-02-2018After sortinguserId: 3, userName: Abhay, createdDate: 11-02-2018userId: 1, userName: Pushpesh, createdDate: 16-02-2018userId: 4, userName: Komal, createdDate: 18-02-2018
侃侃无极
这是javascript中的示例，但可以使用任何类型的字典/地图轻松将其重写为java。people = new Map()user = {id:"das", name: "Darek"}people.set(user.id, user)user = {id:"aaa", name: "Czarrek"}people.set(user.id, user)user = {id:"bbb", name: "Marek"}people.set(user.id, user)console.log(people.get("das"), people.get("aaa"), people.get("bbb"), people.size);people.delete('aaa');console.log(people.get("das"), people.get("aaa"), people.get("bbb"), people.size);映射添加在 O(1) 中工作（对于哈希映射），删除也在 O(1) 中。如果您选择在您的语言中使用的数据结构是使用某种树之王实现的，那么这两个操作都在 O(log(n)) 中运行（可能）。如果只有随机排序的列表并且没有额外的数据，那么唯一的选择就是查看所有列表的项目。它不能比 O(n) 快如果您希望能够快速选择/删除项目，您应该从一开始就使用 Map/Hashmap/Dictionary（在您的语言中如何调用它们无关紧要）（即 O(log n) 或更好）） .