用正确的自动换行将长字符串分成几行

Question

 String original = "This is a sentence.Rajesh want to test the application for the word split.";
 List matchList = new ArrayList();
 Pattern regex = Pattern.compile(".{1,10}(?:\\s|$)", Pattern.DOTALL);
 Matcher regexMatcher = regex.matcher(original);
 while (regexMatcher.find()) {
     matchList.add(regexMatcher.group());
 }
 System.out.println("Match List "+matchList);

我需要将文本解析为长度不超过 10 个字符的行数组，并且在行尾不应有断字。

我在我的场景中使用了以下逻辑，但是如果行尾有中断，它会在 10 个字符后解析到最近的空格

例如：实际的句子是“这是一个句子。Rajesh 想要测试应用程序的单词拆分。”但是在逻辑执行之后它得到如下。

匹配列表[这是一个，nce.Rajesh，想要，测试，应用，对于，单词，拆分。]

Answer 1

好的，所以我已经成功地完成了以下工作，最大行长为 10，但也正确分割了超过 10 的单词！

String original = "This is a sentence. Rajesh want to test the applications for the word split handling.";
List matchList = new ArrayList();
Pattern regex = Pattern.compile("(.{1,10}(?:\\s|$))|(.{0,10})", Pattern.DOTALL);
Matcher regexMatcher = regex.matcher(original);
while (regexMatcher.find()) {
  matchList.add(regexMatcher.group());
}
System.out.println("Match List "+matchList);

这是结果：

This is a 
sentence. 
Rajesh want 
to test 
the 
applicatio
ns word 
split 
handling.

Answer 2

这个问题在某些时候被标记为 Groovy。假设 Groovy 答案仍然有效，并且您不担心保留多个空格（例如 ' '）：

def splitIntoLines(text, maxLineSize) {
    def words = text.split(/\s+/)
    def lines = ['']
    words.each { word ->
        def lastLine = (lines[-1] + ' ' + word).trim()
        if (lastLine.size() <= maxLineSize)
            // Change last line.
            lines[-1] = lastLine
        else
            // Add word as new line.
            lines << word
    }
    lines
}

// Tests...
def original = "This is a sentence. Rajesh want to test the application for the word split."

assert splitIntoLines(original, 10) == [
    "This is a",
    "sentence.",
    "Rajesh",
    "want to",
    "test the",
    "application",
    "for the",
    "word",
    "split."
]
assert splitIntoLines(original, 20) == [
    "This is a sentence.",
    "Rajesh want to test",
    "the application for",
    "the word split."
]
assert splitIntoLines(original, original.size()) == [original]

Answer 3

我避免使用正则表达式，因为它不会减轻重量。此代码自动换行，如果单个单词超过 10 个字符，则将其中断。它还负责处理多余的空白。

import static java.lang.Character.isWhitespace;

public static void main(String[] args) {
  final String original =
    "This is a sentence.Rajesh want to test the application for the word split.";
  final StringBuilder b = new StringBuilder(original.trim());
  final List<String> matchList = new ArrayList<String>();
  while (true) {
    b.delete(0, indexOfFirstNonWsChar(b));
    if (b.length() == 0) break;
    final int splitAt = lastIndexOfWsBeforeIndex(b, 10);
    matchList.add(b.substring(0, splitAt).trim());
    b.delete(0, splitAt);
  }
  System.out.println("Match List "+matchList);
}
static int lastIndexOfWsBeforeIndex(CharSequence s, int i) {
  if (s.length() <= i) return s.length();
  for (int j = i; j > 0; j--) if (isWhitespace(s.charAt(j-1))) return j;
  return i;
}
static int indexOfFirstNonWsChar(CharSequence s) {
  for (int i = 0; i < s.length(); i++) if (!isWhitespace(s.charAt(i))) return i;
  return s.length();
}

打印：

Match List [This is a, sentence.R, ajesh, want to, test the, applicatio, n for the, word, split.]