如何将用户输入拆分为字符数组或字符串数组，然后在java中检查是否有回文

Question

因此，我正在从我的Java编程类编写一个程序，它要求生成一个回文程序。我已经成功地制作了一个程序，该程序只需一个单词的用户输入，就可以运行得很干净，但是我一直在研究如何检查用户单独输入的三个单词。这是到目前为止我的代码。实验室的目标是：逐个检查每个单词，看看它是否是回文——如果你找到了，就把它打印到屏幕上。 我必须遵守这些指示： •要求用户在一行中提供三个回文

•逐个检查每个单词是否为回文–如果找到回文，请打印 它将显示在屏幕上 –提示：查看字符串的toCharArray（）方法 •不断询问用户，直到他们提供了一组至少包含 一个回文

import java.util.Scanner;
public class Palindrome {

private static Scanner scanUserInput;
private static String wordGuess, reverseWord ;
public static void main(String[] args) {
scanUserInput = new Scanner(System.in);
System.out.println("WELCOME TO THE PALINDROME CHECKER!!!");
while(true){
System.out.print("Please enter at least three words to check: ");
wordGuess = scanUserInput.nextLine();
reverseWord = "";

//String[] wordArray = wordGuess.split(","); I tried to use this way to split the inputs but no luck
char[] wordArray = wordGuess.toCharArray();
for(int x=wordArray.length-1;x>=0;x--){
    reverseWord = reverseWord+wordArray[x];

}
System.out.println(reverseWord);
if(wordGuess.equalsIgnoreCase(reverseWord))
{
    System.out.println("");
    System.out.println("You have found a Palindrome!!!");
    System.out.println("The Palindrome we found was "+reverseWord);
    break;
}
else{
    System.out.println("");
    System.out.println("You have not entered a Palindrome...");
    System.out.println("Please Try again...");

}

}//end of main
}
}

谢谢你抽出时间。请给我回任何文件，将有助于我完成这个实验室或任何其他想法。 enter image description here

Answer 1

你的问题是： 实验室的目标是逐个检查每个单词，看看它是否是回文，如果找到了，就把它打印到屏幕上

你必须检查每一个有无回文的字符串，这是我从你的问题中得到的：-

以下是该解决方案的代码：

import java.util.Scanner;
import java.util.StringTokenizer;

/**
 * Created by itt on 10/4/17.
 */
public class Palindrome {
    private static Scanner scanUserInput;
    private static String wordGuess, reverseWord ;
    public static void main(String[] args) {
        scanUserInput = new Scanner(System.in);
        System.out.println("WELCOME TO THE PALINDROME CHECKER!!!");

        System.out.println("Enter Strings with spaces to check palindrome");
        wordGuess = scanUserInput.nextLine();

        StringTokenizer tokens = new StringTokenizer(wordGuess);

        System.out.println("Pallindrome Strings");
        while (tokens.hasMoreTokens()){
            StringBuilder builder = new StringBuilder();
            String token = tokens.nextToken();
            builder.append(token);
            if(builder.reverse().toString().equals(token)) {
                System.out.println(token);
            }
        }

    }
}

输出：

WELCOME TO THE PALINDROME CHECKER!!!
Enter Strings with spaces to check palindrome
abba acac daad jaaj
Pallindrome Strings
abba
daad
jaaj

如果我错了，请纠正我，这样我可以改进

Answer 2

你可以使用split来实现这个目的，它会起作用的

String[] wordArray = wordGuess.split(" "); //or .split("<seperator used b/w words>")

现在迭代单词数组，检查每个单词的回文。 在你的代码中，你没有逐个检查每个单词

//错了

for(int x=wordArray.length-1;x>=0;x--){
    reverseWord = reverseWord+wordArray[x];
}

//修改的代码

   while(true){
        System.out.print("Please enter at least three words to check: ");
        wordGuess = scanUserInput.nextLine();

        String[] wordArray = wordGuess.split(",");
       // char[] wordArray = wordGuess.toCharArray();
        for (String word : wordArray) {
            reverseWord = "";
            for(int x=word.length()-1;x>=0;x--){
                reverseWord = reverseWord+word.charAt(x);

            }
            System.out.println(reverseWord);
            if(word.equalsIgnoreCase(reverseWord))
            {
                System.out.println("");
                System.out.println("You have found a Palindrome!!!");
                System.out.println("The Palindrome we found was "+reverseWord);
            }
            else{
                System.out.println("");
                System.out.println("You have not entered a Palindrome...");
                System.out.println("Please Try again...");

            }
        }


    }

Answer 3

我觉得你的逻辑完全正确。如果您有额外的要求，能够接受CSV单词列表进行检查，那么您可以稍微修改代码，如下所示：

System.out.print("Please enter at least three words to check: ");
String wordGuess = scanUserInput.nextLine();
String[] words = wordGuess.split(",\\s*");
for (String word : words) {
    String reverseWord = "";
    char[] wordArray = word.toCharArray();
    for (int x=wordArray.length-1; x>=0; x--) {
        reverseWord = reverseWord + wordArray[x];
    }
    System.out.println(reverseWord);
   if (wordGuess.equalsIgnoreCase(reverseWord)) {
       System.out.println("");
       System.out.println("You have found a Palindrome!!!");
       System.out.println("The Palindrome we found was " + reverseWord);
    }
    else {
        System.out.println("");
        System.out.println("You have not entered a Palindrome...");
        System.out.println("Please Try again...");
    }
}

我认为你用来分割的图案有问题。你当时是这么做的：

String[] wordArray = wordGuess.split(",");

但是如果你输入一个带有空格的CSV列表，那么它会将空格作为单词的一部分，然后可能不会有回文。此外，你需要循环输入单词，而你也没有这样做

Answer 4

如果用户只输入一个单词，代码就会正常运行。但当你输入多个单词时，till会给出错误。 示例：输入：赛车你好测试 您的代码会将其反转为：tset olleh racecar

赛车你好测试！=tset olleh赛车

如果用户输入的分隔符为，则使用

String[] wordArray = wordGuess.split(",");

将逻辑放在一个单独的函数中：

public checkPalindrome(String word){
    reverseWord="";
    for(int x=word.length-1;x>=0;x--){
        reverseWord = reverseWord+wordArray[x];

    }
    System.out.println(reverseWord);
    if(wordGuess.equalsIgnoreCase(reverseWord))
    {
        System.out.println("");
        System.out.println("You have found a Palindrome!!!");
        System.out.println("The Palindrome we found was "+reverseWord);
        break;
   }
   else{
        System.out.println("");
        System.out.println("You have not entered a Palindrome...");
        System.out.println("Please Try again...");

  }

} 

从main方法调用这个函数三次，并逐个传递所有分隔的字符串

checkPalindrome(wordArray[0]);
checkPalindrome(wordArray[1]);
checkPalindrome(wordArray[2]);

输出：

WELCOME TO THE PALINDROME CHECKER!!!
Enter number of Strings with spaces to check palindrome
abba acac adda abaa
Pallindrome Strings
abba
adda