Question

计算句子中包含正好两个元音的单词数在一对“z”字符之间没有＆＃34; z＆＃34;它们之间。单词之间可能有多个空格，字母可以是大写或小写。显示每个句子中与模式匹配的单词数。

到目前为止，我有以下函数代码for循环，我需要知道如何完成它。请帮帮我。

以下是一些测试用例：

+-------------------------------------------+----------------+
| Sentence                                  | Matching words |
+-------------------------------------------+----------------+
| a ZOO animal is a Zebraz                  | 1              |
| a ZOoZ animal is a Zebraz                 | 2              |
| the zebrazska goozed at the barking zEAlz | 2              |
| azoooza azoo ZORROrroozaster azurez       | 1              |
| nothing much here                         | 0              |
+-------------------------------------------+----------------+

到目前为止，这是我的代码：

private boolean matches(String word)
{
    int Vowels = 3;
    for (int i=3; i<word.length(); ++i) {
        if (word.charAt(i)=='z' )
        {

            return true;
        } else {
        }
    }
    return false;
}

Answer 1

以下是我的意思：

static int countVowels(String atom)
{
  int counter = 0;

  for (int i = 0, l = atom.length(); i < l; i++) {
    char c = atom.charAt(i);

    if ("aeiouAEIOU".indexOf(c) != -1) {
      counter++;
    }
  }

  return counter;
}

static int countMatches(String word)
{
  int count = 0;

  int last_z = -1;

  for (int i = 0, l = word.length(); i < l; i++) {
    char c = word.charAt(i);

    if ((c == 'z') || (c == 'Z')) {
      if (last_z != -1) {
        if (countVowels(word.substring(last_z, (i + 1))) == 2) {
          count++;
        }
      }

      last_z = i;
    }
  }

  return count;
}

public static void main(String[] args)
{
  int total = 0;

  {
    StringTokenizer st = (new StringTokenizer(args[0]));

    while (st.hasMoreTokens()) {
      total += countMatches(st.nextToken());
    }
  }

  System.out.println(total);
}

我如何完成匹配功能/方法？我需要完成这个for循环

1 个答案: