放置 Set<String> 内容的最快方法到一个用空格分隔的单词的字符串? [英] Fastest way to put contents of Set&lt;String&gt; to a single String with words separated by a whitespace?

问题描述

我有几个 Set 并且想将它们中的每一个转换为单个 String，其中原始 Set 由空格"分隔.天真的第一种方法是这样做

I have a few Set<String>s and want to transform each of these into a single String where each element of the original Set is separated by a whitespace " ". A naive first approach is doing it like this

Set<String> set_1;
Set<String> set_2;

StringBuilder builder = new StringBuilder();
for (String str : set_1) {
  builder.append(str).append(" ");
}

this.string_1 = builder.toString();

builder = new StringBuilder();
for (String str : set_2) {
  builder.append(str).append(" ");
}

this.string_2 = builder.toString();

谁能想到更快、更漂亮或更有效的方法来做到这一点?

Can anyone think of a faster, prettier or more efficient way to do this?

推荐答案

使用 commons/lang 你可以使用 StringUtils.join:

String str_1 = StringUtils.join(set_1, " ");

为简洁起见，您真的无法击败它.

You can't really beat that for brevity.

更新:

重新阅读这个答案，我现在更喜欢关于 Guava's Joiner 的另一个答案.事实上，这些天我不去接近 apache commons.

Re-reading this answer, I would prefer the other answer regarding Guava's Joiner now. In fact, these days I don't go near apache commons.

另一个更新:

Java 8 引入了方法 String.join()

Java 8 introduced the method String.join()

String joined = String.join(",", set);

虽然这不像 Guava 版本那样灵活，但当你的类路径上没有 Guava 库时它很方便.

While this isn't as flexible as the Guava version, it's handy when you don't have the Guava library on your classpath.

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