Question

我对JAVA很新。我想为两个版本的回文测试方法中的每个版本绘制两个输入字符串“repaper”和“refinisher”的执行跟踪树

    public static void main( String[] args ) {

        String testCase1 = "repaper";
        String testCase2 = "refinisher";

        if ( isPalindrome_Version_1 ( testCase1 ) )
            System.out.println ( "The string \"" + testCase1 + "\" is a palindrome" );
        else
            System.out.println ( "The string \"" + testCase1 + "\" is NOT a palindrome" );
    }

    // return "true" iff the String s is a palindrome
    private static boolean isPalindrome_Version_1 ( String s )
    {
        boolean result = false;

        // BASE CASE
        if ( s.length() <= 1 )
            return (true);

        // BASE CASE - check if first char equals last char
        if ( s.charAt(0)  !=  s.charAt ( s.length() - 1 ) )
            return ( false );

        // GENERAL CASE - check whether the substring formed by removing 
        //                the first and last letters is a palindrome
        result = isPalindrome_Version_1 ( s.substring ( 1, s.length() - 1 ) );

        return( result );
    }

    // return "true" iff the String s is a palindrome
    private static boolean isPalindrome_Version_2 ( String s )
    {
        boolean result = false;

        // BASE CASE
        if ( s.length() <= 1 )
            return (true);

        // GENERAL CASE - check whether the substring formed by removing 
        //                the first and last letters is a palindrome
        result = isPalindrome_Version_2 ( s.substring ( 1, s.length() - 1 ) );

        // BASE CASE - check if first char equals last char
        if ( s.charAt(0)  !=  s.charAt ( s.length() - 1 ) )
            return ( false );

        return( result  );
    }
}

Answer 1

function Player(username, lvl, exp, gold, hp, atk, def, spd) {
    var self = this;
    this.username = username;
    this.lvl = lvl;
    this.exp = exp;
    this.gold = gold;
    this.hp = hp;
    this.atk = atk;
    this.def = def;
    this.spd = spd;
    this.implement = function() {
        var h1 = document.getElementById('user');
        h1.innerText = this.username;
        h1.addClass('playerName');
        $(h1).data('player', self)
    }
    this.implement();
}
var newPlayer = new Player(prompt("What is your username?"), 1, 0, 0, 10, 2, 2, 2);
playerEl = $('.playerName');
player = playerEl.data('player');

或类似的东西，知道递归的深度

private static boolean isPalindrome_Version_2 ( String s)
{
    boolean result = false;

    // BASE CASE
    if ( s.length() <= 1 ){
         System.out.println("Input Value :" + s + "Return Value" + true);
        return (true);

    // GENERAL CASE - check whether the substring formed by removing 
    //                the first and last letters is a palindrome
    result = isPalindrome_Version_2 ( s.substring ( 1, s.length() - 1 ) );

    // BASE CASE - check if first char equals last char
    if ( s.charAt(0)  !=  s.charAt ( s.length() - 1 ) )
   {
        System.out.println("Input Value :" + s + "Return Value" + false);
        return ( false );
   }

    System.out.println("Input Value :" + s + "Return Value" + result);
    return( result  );
}

顺便说一下，你有另一个递归的解决方案

private static boolean isPalindrome_Version_2 ( String s, int level)
{
    boolean result = false;

    // BASE CASE
    if ( s.length() <= 1 ){
         System.out.println("Input Value :" + s + "Return Value" + true + "Recursion Level" + level);
        return (true);

    // GENERAL CASE - check whether the substring formed by removing 
    //                the first and last letters is a palindrome
    result = isPalindrome_Version_2 ( s.substring ( 1, s.length() - 1 ), level ++ );

    // BASE CASE - check if first char equals last char
    if ( s.charAt(0)  !=  s.charAt ( s.length() - 1 ) )
   {
        System.out.println("Input Value :" + s + "Return Value" + false+ "Recursion Level" + level);
        return ( false );
   }

    System.out.println("Input Value :" + s + "Return Value" + result+ "Recursion Level" + level);
    return( result  );
}

Answer 2

我不认为第2版是正确的（或者至少是有效的）因为你在递归整个字符串后返回false ...

话虽如此，这是第1版的痕迹

isPalindrome("repaper") // length greater than 1 and ending characters are equal
    isPalindrome("epape") // length greater than 1 and ending characters are equal
        isPalindrome("pap") // length greater than 1 and ending characters are equal
            isPalindrome("a") // length is 1, return true

测试案例2

isPalindrome("refinisher") // length greater than 1 and ending characters are equal
    isPalindrome("efinishe") // length greater than 1 and ending characters are equal
        isPalindrome("finish") // length greater than 1 but ending characters are not equal, return false

你现在应该能够自己做第2版

我需要帮助为这个递归方法绘制执行跟踪树

2 个答案: