如何在不迭代的情况下获取给定 LinkedHashSet 元素的索引? [英] How to obtain index of a given LinkedHashSet element without iteration?

问题描述

有可能吗?

说你有

private Set<String> names = new LinkedHashSet<String>();

和 Strings 是M​​ike"、John"、Karen".

and Strings are "Mike", "John", "Karen".

是否有可能在不迭代的情况下得到1"来返回John"的索引是多少?

Is it possible to get "1" in return to "what's the index of "John" without iteration?

以下工作正常..这个问题我想知道是否有更好的方法

The following works fine .. with this question i wonder if there is a better way

for (String s : names) {
    ++i;
    if (s.equals(someRandomInputString)) {
        break;
    }
}

推荐答案

Set interface 没有类似 indexOf() 方法.您确实需要对其进行迭代或使用 List 接口 而是提供 indexOf() 方法.

The Set interface doesn't have something like as an indexOf() method. You'd really need to iterate over it or to use the List interface instead which offers an indexOf() method.

如果您愿意，将 Set 转换为 List 非常简单，只需通过构造函数传递 SetList 的实现.例如

If you would like to, converting Set to List is pretty trivial, it should be a matter of passing the Set through the constructor of the List implementation. E.g.

List<String> nameList = new ArrayList<String>(nameSet);
// ...

这篇关于如何在不迭代的情况下获取给定 LinkedHashSet 元素的索引?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！