Question

我已经看到了其他的递归回文问题而且我已经看到了答案，但我得到的答案有点不同。它需要能够检查带有空格和标点符号的字符串并忽略它们（我知道，它们在技术上不是回文），所以像“女士，我是亚当”这样的字符串应该返回true，但我的程序不是。这就是我所拥有的：

public static boolean isPalindrom(String s){
        System.out.println(s);
        if (!Character.isLetter(s.charAt(0))){
            isPalindrom(s.substring(1,s.length()));
        }
         if (!Character.isLetter(s.charAt(s.length()-1))){
            isPalindrom(s.substring(0,s.length()-1));
        }
         if (s.length() == 1){

            return true;
        }
         if (s.length() == 2){
            if (s.substring(0,1).equalsIgnoreCase(s.substring(s.length()-1))){

                return true;
            }
            else
                return false;
        }
        if (!(s.substring(0,1).equalsIgnoreCase(s.substring(s.length()-1)))){

            return false;
        }

            return isPalindrom(s.substring(1,s.length()-1));

    }

问题是，一旦它开始展开递归，包含空格和标点符号的字符串就开始返回false。我不知道该怎么做。我现在一直在努力尝试不同的解决方案大约一个小时。

P.S。我试图不使用正则表达式来删除空格和标点符号等。

Answer 1

你没有返回前两个isPalindrom的结果（检查现在开始失败的案例......）：

  if (!Character.isLetter(s.charAt(0))){
        return isPalindrom(s.substring(1,s.length()));
    }
     if (!Character.isLetter(s.charAt(s.length()-1))){
        return isPalindrom(s.substring(0,s.length()-1));
    }

Answer 2

当你到达一个时，你只需要继续前进。像这样。我留给你填写空白（特殊情况，lastNonIgnoredChar等）

char[] ignored = new char[] { ',' , ' ', '.'};

int firstNonIgnoredChar(String s) {
    for(int i = 0; i < s.length(); i++) {
        boolean found = false;
        for(char c : ignored) {
            if( c == s.charAt(i) ) found = true;
        }
        if(!found) return i;
    }
    return -1; // no good characters
}

boolean isPal(String s) {
    int first = firstNonIgnoredChar(s);
    int last = lastNonIngoredChar(s);

    if(s.length() == 0) return true;
    if(s.length() == 1) return true;

    return first < last
        && s.charAt(first) == s.charAt(last)
        && isPal(s.substring(first + 1, last - 1);
}

Answer 3

在最坏情况下（完全搜索）在O（n / 2）下运行的非递归方法。当String是LONG的时候，这是更高效的......这是实现......

class PalindromeClass {

    /**
     * This method will run under or at O(n/2) with n = sentence.size()
     * @param sentence is a given String sentence.
     * @return true if the given sentence is a palindrome.
     */
    public static boolean isPalindrome(String sentence) {
        sentence = sentence.replaceAll("[^a-zA-Z0-9]","").toLowerCase();
        char[] sentenceChars = sentence.toCharArray();
        for (int i = 0; i < sentenceChars.length / 2; i++) {
            if (sentenceChars[i] != sentenceChars[sentenceChars.length - 1 - i]) {
                return false;
            }
        }
        return true;
    }

使用short，long和错误参数运行它会为您提供正确的值...

public static void main(String[] args) {

String wrong = "ABCA";
System.out.println("Is '" + wrong + "' a palindrome? ");
System.out.println(isPalindrome(wrong));

String none = "A'";
System.out.println("Is '" + none + "' a palindrome? ");
System.out.print(isPalindrome(none));

String a = "Madam, I'm Adam";
System.out.println("Is '" + a + "' a palindrome? ");
System.out.println(isPalindrome(a));

String b = "abba";
System.out.println("Is '" + b + "' a palindrome? ");
System.out.println(isPalindrome(b));

String toyota = "A Toyota. Race fast, safe car. A Toyota.";
System.out.println("Is '" + toyota + "' a palindrome? ");
System.out.println(isPalindrome(toyota));

String longestPalindrome = "Do good? I? No! Evil anon I deliver. I maim " +
    "nine more hero-men in Saginaw, sanitary sword a-tuck, Carol, I -- lo! " +
    "-- rack, cut a drowsy rat in Aswan. I gas nine more hero-men in Miami. " +
    "Reviled, I (Nona) live on. I do, O God!";
System.out.println("Is '" + longestPalindrome + "' a palindrome? ");
System.out.println(isPalindrome(longestPalindrome));
}

}

这是执行的输出......

Is 'ABCA' a palindrome? false
Is 'a' a palindrome? true
Is 'Madam, I'm Adam a palindrome?' true
Is 'abba a palindrome? true'
Is 'A Toyota. Race fast, safe car. A Toyota.' a palindrome? true
Is 'Do good? I? No! Evil anon I deliver. I maim nine more hero-men in Saginaw, sanitary sword a-tuck, Carol, I -- lo! -- rack, cut a drowsy rat in Aswan. I gas nine more hero-men in Miami. Reviled, I (Nona) live on. I do, O God!' a palindrome? true

需要帮助Java中的palindrome递归方法

3 个答案: