是否有比string.split()更有效的方法将字符串拆分为单词?

你在特定歌曲中搜索一些词吗?如果是这样的话,你可能不需要一套,你可以从你得到歌词的那一点开始搜索。为此,可以使用普通regexp,这可能比拆分字符串、将其放入集合并查询集合快一点,然后:

public class RegexpExample {

public static void main(String[] args) {
    String song = "Is this a real life? Is this just fantasy?";
    String toFind = "is";

    Pattern p = Pattern.compile(toFind, Pattern.CASE_INSENSITIVE);
    Matcher m = p.matcher(song);

    while (m.find()) {
        String found = m.group();
        int startIndex = m.start();
        int endIndex = m.end();

        System.out.println(found + " at start " + startIndex + ", end " + endIndex);
        //do something with this info...
    }
}

它将输出:

Is at start 0, end 2
is at start 5, end 7
Is at start 21, end 23
is at start 26, end 28

但是,如果您搜索不同的歌曲,则可以使用连接它们的歌词 StringBuilder ,然后呼叫 StringBuilder#toString 做整个手术的结果是 toString 方法

有没有特定的方法可以将单词添加到没有 创建数组的中间步骤?

当然,您可以编写一个返回 Iterator 对象,一次输出一个单词。

但像这样的东西确实不值得优化。您的数组很容易就小到可以放入内存中,它的创建成本不会那么高,垃圾收集器随后将对其进行清理。

StringTokenizer st = new StringTokenizer("the days go on and on without you here");
HashSet<String> words = new HashSet<String>();
while (st.hasMoreTokens()) {
    words.add(st.nextToken());
}

我不知道效率,但你也可以这样做:

import java.io.StringReader;

// ...

public static Set<String> getLyricSet(String lyrics) throws IOException {
    StringReader sr = new StringReader(lyrics);
    StringBuilder sb = new StringBuilder();
    Set<String> set = new HashSet<String>();
    int current;
    // Read characters one by one, returns -1 when we're done
    while ((current = sr.read()) != -1) {
        if (Character.isWhitespace(current)) {
            // End of word, add current word to set.
            set.add(sb.toString());
            sb = new StringBuilder();
        } else {
            sb.append((char) current);
        }
    }
    // End of lyrics, add current word to set.
    set.add(sb.toString());
    sr.close();

    return set;
}