小编Rob*_*ley的帖子

所以我在java中制作了一个回文检查器,我似乎遇到了障碍,这是我的代码到目前为止:

public class StringUtil
{

public static void main(String[] args)
{
    System.out.println("Welcome to String Util.");       
    Scanner word = new Scanner(System.in);

    String X = word.nextLine();       
    String R = palindrome(X);

    System.out.println();
    System.out.println("Original Word: " + X);     
    System.out.println("Palindrome: " + R);



}


public static boolean palindrome(String word)
{
   int t = word.length(); //length of the word as a number
   int r = 0;


   if(word.charAt(t) == word.charAt(r))
   {
       return true;
    }
    else
    return false;
}

到目前为止我只想检查第一个字母是否与最后一个字母相同,但是当我编译它时,我在"String R = palindrome(X);"上得到了一个不兼容的类型错误.如何让它在下面的输出语句中打印true或false？

我想知道为什么这不起作用？我已经看了一会儿,想知道为什么它不起作用.它似乎没有读取if语句.即使你输入像"A5"这样的合法字符串,一切看起来都对我不对.

    System.out.println("PLEASE TYPE IN A COORDINATE, LETTER FIRST, THEN NUMBER.");
    System.out.println();
    System.out.print("CAPITALS ONLY, PLEASE. ENTER COORDINATE HERE: ");

    String placement = scan.next();
    String letter = placement.substring(0,1);
    String number = placement.substring(1);

    System.out.println(letter);
    //System.out.println(placement);
    System.out.println(number);

    String cord[] = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J"};

    for(int s = 0; s < cord.length; s++)
    {
        //System.out.println(cord[s]);
        if(cord[s] == letter)
        {
            System.out.print("yay, it worked");
        }
    }