英文:
Java class for storing objects by index
问题
抱歉，不知道如何在标题中简洁地解释。我需要一个存储<对象编号，对象>对并自动生成递增编号的Java类。如果删除了某个项目，则其编号应该重新用于下一个存储的项目。就像这样：
public class SerialStorage<T>
{
Map<Integer, T> itemsStorage;
Queue<Integer> deletedItemsStorage;
int lastKnownMaxNumber;
public SerialStorage()
{
itemsStorage = new HashMap<>();
deletedItemsStorage = new ArrayDeque<>();
lastKnownMaxNumber = 0;
}
/*
存储新项目并返回其编号
*/
public int putItem(T item)
{
int number;
if (deletedItemsStorage.size() > 0)
{
number = deletedItemsStorage.remove();
}
else
{
number = ++lastKnownMaxNumber;
}
itemsStorage.put(number, item);
return number;
}
/*
根据编号获取项目
*/
public T getItem(int number)
{
return itemsStorage.get(number);
}
/*
删除项目并返回它。如果找不到此项目，则返回null
*/
public T removeItem(int number)
{
T removedItem = itemsStorage.get(number);
if (removedItem != null)
{
deletedItemsStorage.add(number);
}
return removedItem;
}
}
Java标准库中是否有类似的内容？
英文:
Sorry, don't know how to explain it in brief in the title.
I need a java class that stores <object number, object> pairs generating incremental numbers automatically. If some item is deleted, the number it has shall be reused for the next stored item. Like that:
public class SerialStorage<T>
{
Map<Integer, T> itemsStorage;
Queue<Integer> deletedItemsStorage;
int lastKnownMaxNumber;
public SerialStorage()
{
itemsStorage = new HashMap<>();
deletedItemsStorage = new ArrayDeque<>();
lastKnownMaxNumber = 0;
}
/*
Stores new item and returns it's number
*/
public int putItem(T item)
{
int number;
if (deletedItemsStorage.size() > 0)
{
number = deletedItemsStorage.remove();
}
else
{
number = ++lastKnownMaxNumber;
}
itemsStorage.put(number, item);
return (number);
}
/*
Get item by it's number
*/
public T getItem(int number)
{
return (itemsStorage.get(number));
}
/*
Removes item and returns it. Returns null if no such item found
*/
public T removeItem(int number)
{
T removedItem = itemsStorage.get(number);
if (removedItem != null)
{
deletedItemsStorage.add(number);
}
return (removedItem);
}
}
Is there anything like this in the java standard libraries?
答案1
得分: 0
我猜你在寻找一个ArrayList？
它们会自动编索引，当你从中间移除一个项目时，所有后续的项目将会向下滑动一个索引。
一个简单的示例：
ArrayList<String> myStrings = new ArrayList<String>();
strings.add("this is a test"); // 索引 0
strings.add("this is number two"); // 索引 1
strings.add("this is number three"); // 索引 2
System.out.println(strings.get(1));
strings.remove(1);
System.out.println(strings.get(1));
在此之后，索引 0 处将会是 "this is a test"，索引 1 处将会是 "this is number three"。
英文:
I guess you´re looking for an ArrayList ?
They auto-index and they clean up so if you remove an item from the middle all the following items will slide one index down.
A brief example:
ArrayList<String> myStrings = new ArrayList<String>();
strings.add("this is a test"); // index 0
strings.add("this is number two"); // index 1
strings.add("this is number three"); // index 2
System.out.println(strings.get(1));
strings.remove(1);
System.out.println(strings.get(1));
After this you´ll have "this is a test" at index 0 and "this is number three" at index 1.
答案2
得分: 0
你可能需要的是一个映射（Map）。但它不会自动生成索引，这由你来完成。
Map<Integer, String> map = new HashMap<Integer, String>();
map.put(findMissingPositive(map.keySet(), map.keySet().size()), "New String");
你可以使用一个函数来获取键中缺失的最小数字。
static int findMissingPositive(int arr[], int size)
{
int i;
// 通过将 arr[i] 设为负数来标记 arr[i] 为已访问。注意，需要减去 1，因为索引从 0 开始，而正数从 1 开始。
for (i = 0; i < size; i++) {
int x = Math.abs(arr[i]);
if (x - 1 < size && arr[x - 1] > 0)
arr[x - 1] = -arr[x - 1];
}
// 返回第一个正数的索引值
for (i = 0; i < size; i++)
if (arr[i] > 0)
return i + 1; // 需要加上 1，因为索引从 0 开始
return size + 1;
}
https://www.geeksforgeeks.org/find-the-smallest-positive-number-missing-from-an-unsorted-array/
英文:
What you might need is a Map. But it won't generate the indexes automatically That is up to you.
Map<Integer, String> map = New HashMap<Integer,String>();
map.add(findMissingPositive(map.KeySet(), map.KeySet().size()), "New String");
You might use a function that retrieves the smallest number that is missing from the keys.
static int findMissingPositive(int arr[], int size)
{
int i;
// Mark arr[i] as visited by making
// arr[arr[i] - 1] negative. Note that
// 1 is subtracted because index start
// from 0 and positive numbers start from 1
for (i = 0; i < size; i++) {
int x = Math.abs(arr[i]);
if (x - 1 < size && arr[x - 1] > 0)
arr[x - 1] = -arr[x - 1];
}
// Return the first index value at which
// is positive
for (i = 0; i < size; i++)
if (arr[i] > 0)
return i + 1; // 1 is added becuase indexes
// start from 0
return size + 1;
}
https://www.geeksforgeeks.org/find-the-smallest-positive-number-missing-from-an-unsorted-array/
专注分享java语言的经验与见解，让所有开发者获益！
评论