Question

所以我最近上大学并且表现相当不错，这导致我加入了编程竞赛。

在一项任务中，我们被要求创建一个程序，要求用户输入一个字符串。之后，程序需要查看字符串是否为回文结构而忽略空格和特殊字符。

实施例：马...... a.m：回文没有....魔鬼......，.'....生活......在

即使在比赛结束后，我也尽力回答这个问题。我设法通过它并且有资格参加下一轮比赛，但这项任务让我在夜间做了噩梦lol

Answer 1

我会像这个伪代码一样解决这个问题：

letters <== ""

characterEntered ( letters, c ) {
    if (c is an uppercase letter) lowercase(c)
    if (c is a letter) append c to letters
    /* otherwise forget about it */
}

allCharactersEntered ( list ) {
    backwards <== reverse(letters)
    if (backwards equals letters) say "Palindrome"
    else say "Nope"
}

Answer 2

Java版：

public static void main(String[] args) {

    palin();
}

public static void palin() {

    String in = null;
    // loop
    while (true) {
        // read input
        in = readInput();
        // check input
        if ("q".equalsIgnoreCase(in)) {
            // exit
            break;
        }
        if (checkPalindrome(in)) {
            System.out.println("Pass");
        } else {
            System.out.println("Fail");
        }

    }

}

private static boolean checkPalindrome(String in) {
    String sanitizedIn = sanitize(in);

    return sanitizedIn.equalsIgnoreCase(StringUtils.reverse(sanitizedIn));
}

private static String sanitize(String in) {
    String result = null;

    Pattern p = Pattern.compile("[^a-zA-Z0-9]+");
    Matcher m = p.matcher(in);
    result = m.replaceAll("");
    System.out.println(in + "|>>>>>|" + result);
    return result;
}

private static String readInput() {

    try {
        System.out.println("Enter text(q to Quit):");
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        return reader.readLine();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
        return null;
    }
}

}

Answer 3

这是更有效的解决方案

private boolean isPalindrome(String text) {
    int index1 = 0;
    int index2 = text.length() - 1;
    char[] chars = text.toCharArray();
    if (!Character.isLetter(chars[index1])) {
      index1 = nextLetterIndex(chars, index1);
    }
    if (!Character.isLetter(chars[index2])) {
      index2 = previousLetterIndex(chars, index2);
    }
    while (index2 >= index1) {
      if (Character.toLowerCase(chars[index1]) != Character.toLowerCase(chars[index2])) {
        return false;
      }

      index1 = nextLetterIndex(chars, index1);
      index2 = previousLetterIndex(chars, index2);
    }

    return true;
  }

  int previousLetterIndex(char[] chars, int index) {
    index--;
    while (!Character.isLetter(chars[index])) {
      index--;
    }
    return index;
  }

  int nextLetterIndex(char[] chars, int index) {
    index++;
    while (!Character.isLetter(chars[index])) {
      index++;
    }
    return index;
  }

Answer 4

public boolean isPalindromeSpecialCharacters(String input)
{
    if (input == null)
        return false;

    String stringWithoutSpecials = "";

    char charArr[] = input.toCharArray();

    for (char ch : charArr)
    {
        if (Character.isLetter(ch))
            stringWithoutSpecials += ch;
    }

    if (stringWithoutSpecials.equals(this.reverseString(stringWithoutSpecials)))
        return true;

    return false;
}

public String reverseString(String text)
{
    if (text == null || text.isEmpty())
        return text;

    return reverseString(text.substring(1)) + text.charAt(0);
}

回文：忽略空格，案例，特殊字符

4 个答案: