問(wèn)題描述

我有一組對(duì)象.每個(gè)對(duì)象都有 String 值.

I have Set of objects. Each object has String value.

我需要選擇所有 this 值等于direction"的對(duì)象.

I need to select all objects that have this value equal to "direction".

是否可以不迭代集合?

推薦答案

一般來(lái)說(shuō)，沒(méi)有.您需要遍歷集合并檢查每個(gè)對(duì)象以查看屬性是否等于您正在搜索的值.這是一個(gè) O(n) 操作.

In general, no. You need to iterate over the set and check each object to see if the property is equal to the value you are searching for. This is an O(n) operation.

在一種情況下，您無(wú)需迭代即可完成.如果您的對(duì)象的 equals 方法是根據(jù)該 String 屬性的相等性定義的，并且如果 hashCode 方法也正確實(shí)現(xiàn)，那么您可以使用 hashSet.contains 在 O(1) 時(shí)間內(nèi)找到具有正確值的對(duì)象，而無(wú)需遍歷集合.

There is one situation in which you could do it without iterating. If your object's equals method is defined in terms of equality of that String property, and if the hashCode method is also implemented correctly, then you can use the hashSet.contains to find an object with the correct value in O(1) time without requiring iterating over the set.

正如我所提到的，這是一個(gè)非常具體的用例，而不是通用解決方案.如果字符串是某種唯一標(biāo)識(shí)符，它可能會(huì)很有用，但它不適用于您的特定用例.

As I mentioned, this is a very specific use case and not a general solution. It might be useful if the string was some sort of unique identifier, but it won't work for your specific use case.

您可能還想考慮其他更適合您的用例的集合.例如，如果您使用 Guava，那么您可以考慮使用 多地圖.

You might also want to consider other collections that would be better suited to your use case. You could for example if you are using Guava then you could consider using a Multimap.

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