字符串中最长回文的输出将被正确打印，尽管它不是堆栈的一部分。

Question

我编写了一些代码来查找字符串中最长的回文(回文不必出现在一起，因为它可以是不连续的)。

它适用于几乎所有的案件。对于下面代码中的情况，它也会打印出正确的回文及其长度。然而，有一个问题使我感到困惑。我有一个名为compare()的函数，在这里，我将一个新发现的回文的长度与'longestPalindromeLength‘进行比较，其思想是，当所有助手函数返回main时，名为'longestPalindromeString’的静态(全局)变量将得到结果。

我的问题是，当我打印它时，我没有看到最长的回文，它是"ABCDEEEEDCBA“在这个比较法()函数中的任何地方。

请看我的密码

public class LongestPalindromeNonContiguousPrint
{
    //static String S = "abcdcba";
    //static String S = "SGEGGES";
    static String S = "SGEGGESABCDEEEEDCBA";
    //static String S = "abca1221";

    static int longestPalindromeLength = 0;
    static String longestPalindromeString = "";

    public static void main(String[] args)
    {
        System.out.println("Length of the longest palindrome == " + fun(0, S.length()-1,""));
        System.out.println("Longest palindrome == "+longestPalindromeString);
    }

    static int fun(int s, int e, String palindrome)
    {
        String temp = "";

        /* base cases for even */
        if(s == e-1)
        {
            if(S.charAt(s) == S.charAt(e))
            {
                palindrome = palindrome + S.charAt(s);
                compare(palindrome,"even");
                return 2;
            }

            else
            {
                palindrome = palindrome + S.charAt(s);
                compare(palindrome,"odd");
                return 1;
            }
        }

        /* base case for odd */
        if(s == e)
        {
            palindrome = palindrome + S.charAt(s);
            compare(palindrome,"odd");
            return 1;
        }

        /*if(s > e)
            return (S.charAt(s-1) == S.charAt(e+1)) ? 1:0;*/

        /* recurse */
        if(S.charAt(s) == S.charAt(e))
        {
            palindrome = palindrome + S.charAt(s);

            temp = palindrome;
            int rec = fun(s+1, e-1, palindrome);
            palindrome = temp;

            int ret = 2 + rec;
            return ret;
        }

        else 
        {
            temp = palindrome;
            int rec1 = fun(s+1, e, palindrome);
            palindrome = temp;

            temp = palindrome;
            int rec2 = fun(s, e-1, palindrome);
            palindrome = temp;

            return max(rec1, rec2);
        }
    }

    static int max(int a, int b)
    {
        if(a > b)
            return a;
        return b;
    }

    static void compare(String s, String type)
    {
        String palindrome = "";
        String rev = new StringBuilder(s).reverse().toString();

        if(type == "odd")
        {
            palindrome = s + rev.substring(1,rev.length());
        }

        else if(type == "even")
        {
            palindrome =  s + rev;
        }

        if(palindrome.length() > longestPalindromeLength)
        {
            longestPalindromeLength = palindrome.length();
            longestPalindromeString = palindrome;

            /* This does not get printed, I do not understand where this print() function
             * sees this string ABCDEEEEDCBA */
            if(longestPalindromeString == "ABCDEEEEDCBA")
            {
                System.out.println("found ABCDEEEEDCBA");
            }
        } 
    }
}

输出

Length of the longest palindrome == 12
Longest palindrome == ABCDEEEEDCBA

请看一下compare()函数，我插入了一个if-条件来打印"ABCDEEEEDCBA“，而这是最长的回文。但它从来没有碰到过这种情况。

编辑:如果输出太大，eclipse是否会修剪掉一些输出。对于下面的程序，我观察到eclipse和从终端运行的输出之间的差异。运行在eclipse上的输出为24811行，而从终端运行的输出为47769行。

public class LongestPalindromeNonContiguousPrint
{
    //static String S = "abcdcba";
    //static String S = "GEEKSFORGEEKS";
    //static String S = "SGEGGES";
    static String S = "SGEGGESABCDEEEEDCBA";
    //static String S = "abca1221";

    static int longestPalindromeLength = 0;
    static String longestPalindromeString = "";

    public static void main(String[] args)
    {
        System.out.println("Length of the longest palindrome == " + fun(0, S.length()-1,""));
        System.out.println("Longest palindrome == "+longestPalindromeString);
    }

    static int fun(int s, int e, String palindrome)
    {
        String temp = "";

        /* base cases for even */
        if(s == e-1)
        {
            if(S.charAt(s) == S.charAt(e))
            {
                palindrome = palindrome + S.charAt(s);
                compare(palindrome,"even");
                return 2;
            }

            else
            {
                palindrome = palindrome + S.charAt(s);
                compare(palindrome,"odd");
                return 1;
            }
        }

        /* base case for odd */
        if(s == e)
        {
            palindrome = palindrome + S.charAt(s);
            compare(palindrome,"odd");
            return 1;
        }

        /*if(s > e)
            return (S.charAt(s-1) == S.charAt(e+1)) ? 1:0;*/

        /* recurse */
        if(S.charAt(s) == S.charAt(e))
        {
            palindrome = palindrome + S.charAt(s);

            temp = palindrome;
            int rec = fun(s+1, e-1, palindrome);
            palindrome = temp;

            int ret = 2 + rec;
            return ret;
        }

        else 
        {
            temp = palindrome;
            int rec1 = fun(s+1, e, palindrome);
            palindrome = temp;

            temp = palindrome;
            int rec2 = fun(s, e-1, palindrome);
            palindrome = temp;

            return max(rec1, rec2);
        }
    }

    static int max(int a, int b)
    {
        if(a > b)
            return a;
        return b;
    }

    static void compare(String s, String type)
    {
        String palindrome = "";
        String rev = new StringBuilder(s).reverse().toString();

        if(type == "odd")
        {
            palindrome = s + rev.substring(1,rev.length());
        }

        else if(type == "even")
        {
            palindrome =  s + rev;
        }

        System.out.println(palindrome);

        if(palindrome.length() > longestPalindromeLength)
        {
            longestPalindromeLength = palindrome.length();
            longestPalindromeString = palindrome;

            /*if(palindrome.equals("ABCDEEEEDCBA"))
            {
                System.out.println("found ABCDEEEEDCBA");
            }*/
        } 
    }
}

Answer 1

您正在将字符串与==而不是String的equals方法进行比较。