使用substring（）和charAt（）檢查String是否為Palindrome

Question

我的代碼應該要求用戶輸入一個String，然后使用charAt()和substring()方法檢查它是否是Palindrome。 我相信，我已經做了一切正確但NetBeans正在編譯錯誤。

當我運行我的代碼時，NetBeans告訴我：

Exception in thread "main" java.lang.RuntimeException: Uncompilable source code - missing return statement
    at palindrome.Palindrome.methodA(Palindrome.java:31)
    at palindrome.Palindrome.main(Palindrome.java:27)
Java Result: 1

所以，基本上它說我錯過了一個返回聲明？ 我不明白這是怎么可能的。

任何幫助將非常感謝伙計們。 謝謝。

package palindrome;

import java.util.Scanner;

public class Palindrome {

    /**
     * Program asks user to enter a string.
     */
    public static void main(String[] args) {
        String userInput; //user-inputted String
        String result;

        //set up instance of Scanner for user input
        Scanner scan = new Scanner(System.in);

        //ask user for input
        System.out.print("Please enter a String: ");
        userInput = scan.nextLine();

        //run methodA
        result = methodA(userInput);

    }

    public static String methodA(String inString) {
        /*
         * This method checks the user-inputted String against a backward copy
         * of itself using only String and Character methods.
         */

        //next two int variables used as int pointers for first letter of String
        //and last letter of user-inputted String
        int begPoint = 0;
        int endPoint = inString.length() - 1;

        //define Strings to return to main method
        String isPal = (inString + " is a palindrome!");
        String notPal = (inString + " is not a palindrome.");

        while (begPoint < endPoint) {
            //two substrings defined to test String character for character
            String firstChar = inString.substring(begPoint, begPoint + 1);
            String lastChar = inString.substring(endPoint, endPoint + 1);

            //algorithm continues step by step with begPoint going up one and 
            //endPoint decreasing by one
            begPoint++;
            endPoint--;

            //basically here I am trying to check the characters of the user-
            //inputted Strings using the charAt methods but I'm lost when I
            //get to this point
            if (inString.charAt(begPoint) == inString.charAt(endPoint)) {
                return isPal;
            } else {
                return notPal;
            }
        }
    }
}

經過一些回答 - 我現在已將我的while語句更改為：

while(begPoint<endPoint) {
    //two substrings defined to test String character for character
    String firstChar = inString.substring(begPoint, begPoint + 1);
    String lastChar = inString.substring(endPoint, endPoint + 1);

    //basically here I am trying to check the characters of the user-
    //inputted Strings using the charAt methods 
    if (inString.charAt(begPoint) == inString.charAt(endPoint)) {
        return isPal;
    } else {
        return notPal;
    }

    //algorithm continues step by step with begPoint going up one and 
    //endPoint decreasing by one
    begPoint++;
    endPoint--;
}

但是，現在它告訴我（begPoint ++）是一個無法訪問的語句。

Answer 1

在這段代碼中

if(inString.charAt(begPoint)==inString.charAt(endPoint))
{
       return isPal;
}
else
{
       return notPal;
} 

你基本上是在循環的第一次迭代中返回。

如果它不是Palindrome，那么立即返回是有意義的，但是其他方面你需要繼續循環。

如果沒有輸入while ，代碼也不會編譯。 也許返回notPal

另外（感謝ArtOfWarfare），你需要搬家

//algorithm continues step by step with begPoint going up one and 
//endPoint decreasing by one
    begPoint++;
    endPoint--;

到你的while循環結束，即在完成if語句之后。

Answer 2

你的問題是你在檢查charAts之前移動了begPoint和endPoint。 在使用方法charAt之后，將代碼前進begPoint和endPoint移動到。