java我想在特定长度后拆分一个字符串，而不剪切任何单词，而不是相等的字符串

Question

我想把一根绳子分开，比如“贰佰肆拾壹元陆角捌卢比整”，当它的长度是35时，我想把这根绳子分成两半。我尝试了这个代码来分割字符串

String text = "Rupees Two Hundred Forty One and Sixty Eight only";
List<String> parts = new ArrayList<>();
int length = text.length();
for (int i = 0; i < length; i += 35) {
    parts.add(text.substring(i, Math.min(length, i + size)));

但是输出是这样的

[Rupees Two Hundred Forty One and Si, xty Eight only]

但我想像这样分开绳子

[Rupees Two Hundred Forty One and, Sixty Eight only]

拆分字符串时没有剪切词。 根据账单金额，字符串每次都会变化

Answer 1

只需在i+35位置搜索首选位置。要考虑的一件事是，当没有这样的位置，即单词超过指定的大小时，应该发生什么。下面的代码将执行大小限制，如果找不到好的位置，则在单词的中间断裂：

List<String> parts = new ArrayList<>();
int size = 35, length = text.length();
for(int i = 0, end, goodPos; i < length; i = end) {
    end = Math.min(length, i + size);
    goodPos = text.lastIndexOf(' ', end);
    if(goodPos <= i) goodPos = end; else end = goodPos + 1;
    parts.add(text.substring(i, goodPos));
}

如果中断发生在空格字符处，则将从结果字符串中删除空格

Answer 2

您可以找到“and”的索引，并将字符串从0子串到“and”的索引

 int i = text.indexOf("and") + 3;
 String part1 = text.substring(0,i);
 String part2 = text.substring(i).trim();

Answer 3

我将使用StringBuilders从头开始构建字符串。下面是一个例子，里面有一些评论：

    String text = "Rupees Two Hundred Forty One and Sixty Eight only For seven thousand chickens";
    String split[] = text.split(" "); // Split by space
    // One SB for each sentence
    StringBuilder sentence = new StringBuilder();
    // One SB for the total String
    StringBuilder total = new StringBuilder();
    for (int i = 0; i < split.length; i++) {
        String word = split[i];
        // Check if that words fits to sentence
        if (sentence.length() + word.length() <= 35) {
            sentence.append(word);
            sentence.append(" ");
        } else {
            total.append(sentence.toString().trim());
            total.append(", ");
            // Flush sentence to total and start next sentence
            sentence = new StringBuilder();
            sentence.append(word);
            sentence.append(" ");
        }
    }
    //Add any leftover
    if (sentence.length() > 0)
        total.append(sentence.toString().trim());
    System.out.println(total.toString());

哪些输出到：

Rupees Two Hundred Forty One and, Sixty Eight only For seven thousand, chickens

Answer 4

我认为可以使用while循环来计算包含最后一个空格字符的单词：

public static List<String> split(String str, int length) {
    List<String> res = new ArrayList<>();
    int prvSpace = 0;
    int from = 0;

    while (prvSpace < str.length()) {
        int pos = str.indexOf(' ', prvSpace + 1);

        if (pos == -1) {
            res.add(str.substring(from));
            prvSpace = str.length();
        } else if (pos - from < length)
            prvSpace = pos;
        else {
            res.add(str.substring(from, prvSpace));
            from = prvSpace + 1;
        }
    }

    return res;
}

---

演示：

in: "RupeesTwoHundredFortyOneandSixtyEightonly"
out: ["RupeesTwoHundredFortyOneandSixtyEightonly"]

in: "Rupees Two Hundred Forty One and Sixty Eight only"
out: ["Rupees Two Hundred Forty One and", "Sixty Eight only"]

Answer 5

你可能无法完全做到这一点。但是使用字符串。indexOf（）查找从35开始的第一个空格。然后使用substring方法分割字符串

      String text = "Rupees Two Hundred Forty One and Sixty Eight only";
      int i = text.indexOf(" ", 35);
      if (i < 0) {
         i = text.length();
      }
      String part1 = text.substring(0,i).trim();
      String part2 = text.substring(i).trim();

这里有另一种方法。尚未对边境案件进行全面检查

      String[] words = text.split(" ");
      int k;
      part1 = words[0];
      for (k = 1; k < words.length; k++) {
         if (part1.length() >= 35 - words[k].length()) {
            break;
         }
         part1 += " " + words[k];
      }
      if (k < words.length) {
         part2 = words[k++];
         while (k < words.length) {
            part2 += " " + words[k++];
         }
      }
      System.out.println(part1);
      System.out.println(part2);