从Java字符串中以长度1的字符串返回的第一个字母的最佳方法是什么？

Question

假设以下内容：

String example      = "something";
String firstLetter  = "";

以下分配方式firstLetter可能会影响性能，请注意是否存在差异； 哪个最好，为什么？

firstLetter = String.valueOf(example.charAt(0));
firstLetter = Character.toString(example.charAt(0));
firstLetter = example.substring(0, 1);

返回第一个字母为a的原因String是它正在Hadoop中运行，并且需要一个字符串来分配给一个Text类型，firstLetter它将key从map()方法中以a
形式输出，例如：

public class FirstLetterMapper extends Mapper<LongWritable, Text, Text, IntWritable> {
    String line = new String();
    Text firstLetter = new Text();
    IntWritable wordLength = new IntWritable();

    @Override
    public void map(LongWritable key, Text value, Context context)
            throws IOException, InterruptedException {

        line = value.toString();

        for (String word : line.split("\\W+")){
            if (word.length() > 0) {

                // ---------------------------------------------
                // firstLetter assignment
                firstLetter.set(String.valueOf(word.charAt(0)).toLowerCase());
                // ---------------------------------------------

                wordLength.set(word.length());
                context.write(firstLetter, wordLength);
            }
        }
  }
}

Answer 1

substring(0, 1)通过执行以下操作，可以发现性能更好：

    String example = "something";
    String firstLetter  = "";

    long l=System.nanoTime();
    firstLetter = String.valueOf(example.charAt(0));
    System.out.println("String.valueOf: "+ (System.nanoTime()-l));

    l=System.nanoTime();
    firstLetter = Character.toString(example.charAt(0));
    System.out.println("Character.toString: "+ (System.nanoTime()-l));

    l=System.nanoTime();
    firstLetter = example.substring(0, 1);
    System.out.println("substring: "+ (System.nanoTime()-l));

输出：

String.valueOf: 38553
Character.toString: 30451
substring: 8660