blocks|key|3189234|text|您的解决方案至少有两个基本问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3189235|您将弹出s1，直到它为空，然后执行回文检查。|ordered-list-item|3189236|s1和反转器不是相同的数据类型。|3189237|1)暂时忽略项目2，当您使用s1来确定输入是否为回文时，它看起来是空的。您需要确保在进行等效检查时，两个变量都指向包含数据的结构！|3189238|2)+s1是堆栈，反转器是String[]。Stack中的“equals”的默认实现是检查参数是否是堆栈的实例。因为永远不会出现这种情况，所以您将始终看到“输入字符串不是回文”。|3189239|要修复2)，需要使两个对象都具有相同的数据类型(无论是堆栈还是String[])。如果选择后一个String[]，则需要验证每个索引i处的字符串在s1和反转器中都是等效的。|3189240|希望这能有所帮助！|3189241|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$]]|$1|E|3|F|5|D|7|S|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|T|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|U|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|V|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|W|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|P|$]]

There are at least two fundamental problems with your solution:

<ol>
<li>You are popping s1 until it is empty and then doing your palindrome check</li>
<li>s1 and reverser are not the same data type</li>
</ol>

1) Ignoring bullet 2 for the time being, it looks like s1 is empty when you use it to determine if the input is a palindrome. You'll need to make sure that when you do an equivalence check both the variables point to structures with data in them! 

2) s1 is a Stack and reverser is a String[]. The default implementation of 'equals' in Stack is to check if the argument is an instance of a Stack. Since this will never be the case, you will always see "The input String is not a palindrome.". 

To fix 2), you'll need to make both objects the same data type (either both Stack or String[]). If you choose the latter String[], you will need to verify that the Strings at each index i are equivalent in both s1 and reverser.

Hope this helps!

blocks|key|1179846|text|检查堆栈类中等于方法的实现，因为实现没有发布，因此很难找到问题。但是，如果您不需要这样做，您可以使用相同的方法进行比较。记住使用相同的数据类型。例如，将数组和堆栈强制转换为字符串。然后使用string类的相等方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1179847|String+original+=+"level";+//The+contents+of+you+stack+dumped+into+a+String
String[]+reversedArr+=+{"l",+"e",+"v",+"e",+"l"};+//And+your+array+reverser

String+reversed+=+"";

for+(String+letter+:+reversedArr)
++++reversed+=+reversed.concat(letter);

System.out.println(original.equals(reversed));|code-block|syntax|javascript|1179848|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Check the implementation of the equals method in your stack class, since the implementation is not posted is hard to find the problem. But instead if you don't need to do it you can do the comparison in the same method. Remember to use the same data type. For example cast the array and the stack to a string. Then use the equals method of the string class.

<pre><code>String original = "level"; //The contents of you stack dumped into a String
String[] reversedArr = {"l", "e", "v", "e", "l"}; //And your array reverser

String reversed = "";

for (String letter : reversedArr)
 reversed = reversed.concat(letter);

System.out.println(original.equals(reversed));
</code></pre>

blocks|key|3189306|text|我在20年没有使用Java之后，重新学习它。下面是我之前对你的主要帖子的评论的一个实现。我的堆栈是基于OOP的，而不是使用内部数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3189307|public+class+Palindrome+{

++++public+static+void+main(String[]+args)
++++{
++++++++java.util.Scanner+keyboard+=+new+java.util.Scanner(System.in);
++++++++System.out.print("Enter+the+word+to+check:+");
++++++++String+input+=+keyboard.nextLine();
++++++++Palindrome+p+=+new+Palindrome(input);
++++++++System.out.println("Input:+"+%2B+p.getInput());
++++++++System.out.println("Reversed:+"+%2B+p.getInputReversed());
++++++++System.out.println("Palindromic:+"+%2B+p.isPalindromic());
++++}

++++private+String+_input;

++++public+Palindrome(String+input)
++++{
++++++++_input+=+input.toLowerCase();
++++}

++++public+String+getInput()
++++{
++++++++return+_input;
++++}

++++public+String+getInputReversed()
++++{
++++++++CharStack+chars+=+new+CharStack(getInput());++++
++++++++String+reversed+=+"";
++++++++while(!chars.empty())
++++++++{
++++++++++++try+{
++++++++++++++++reversed+=+reversed+%2B+chars.pop();
++++++++++++}+catch+(Exception+e)+{
++++++++++++++++System.out.println("Exception:+"+%2B+e.getMessage());
++++++++++++}
++++++++}
++++++++return+reversed;
++++}

++++public+boolean+isPalindromic()
++++{
++++++++return+getInput().equals(getInputReversed());
++++}

++++private+class+CharStack
++++{

++++++++private+CharStackNode+top+=+null;

++++++++private+class+CharStackNode
++++++++{
++++++++++++public+char+c;
++++++++++++public+CharStackNode+next+=+null;

++++++++++++public+CharStackNode(char+inC,+CharStackNode+top)
++++++++++++{
++++++++++++++++c+=+inC;
++++++++++++++++next+=+top;
++++++++++++}
++++++++}

++++++++public+CharStack(String+input)
++++++++{
++++++++++++for(char+c:+input.toCharArray())
++++++++++++{
++++++++++++++++push(c);
++++++++++++}
++++++++}

++++++++public+void+push(char+c)
++++++++{
++++++++++++top+=+new+CharStackNode(c,+top);;
++++++++}

++++++++public+char+pop()+throws+Exception
++++++++{
++++++++++++if+(!empty())
++++++++++++{
++++++++++++++++char+c+=+top.c;
++++++++++++++++top+=+top.next;
++++++++++++++++return+c;
++++++++++++}
++++++++++++else
++++++++++++{
++++++++++++++++throw+new+Exception("Cannot+pop()+an+empty+stack!");
++++++++++++}
++++++++}

++++++++public+boolean+empty()
++++++++{
++++++++++++return+top+==+null;
++++++++}

++++}

}|code-block|syntax|javascript|3189308|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I'm re-learning Java after not using it for two decades. Here's an implementation of my previous comment on your main post. My stack is OOP based instead of using an internal array:

<pre><code>public class Palindrome {

 public static void main(String[] args)
 {
 java.util.Scanner keyboard = new java.util.Scanner(System.in);
 System.out.print("Enter the word to check: ");
 String input = keyboard.nextLine();
 Palindrome p = new Palindrome(input);
 System.out.println("Input: " + p.getInput());
 System.out.println("Reversed: " + p.getInputReversed());
 System.out.println("Palindromic: " + p.isPalindromic());
 }

 private String _input;

 public Palindrome(String input)
 {
 _input = input.toLowerCase();
 }

 public String getInput()
 {
 return _input;
 }

 public String getInputReversed()
 {
 CharStack chars = new CharStack(getInput()); 
 String reversed = "";
 while(!chars.empty())
 {
 try {
 reversed = reversed + chars.pop();
 } catch (Exception e) {
 System.out.println("Exception: " + e.getMessage());
 }
 }
 return reversed;
 }

 public boolean isPalindromic()
 {
 return getInput().equals(getInputReversed());
 }

 private class CharStack
 {

 private CharStackNode top = null;

 private class CharStackNode
 {
 public char c;
 public CharStackNode next = null;

 public CharStackNode(char inC, CharStackNode top)
 {
 c = inC;
 next = top;
 }
 }

 public CharStack(String input)
 {
 for(char c: input.toCharArray())
 {
 push(c);
 }
 }

 public void push(char c)
 {
 top = new CharStackNode(c, top);;
 }

 public char pop() throws Exception
 {
 if (!empty())
 {
 char c = top.c;
 top = top.next;
 return c;
 }
 else
 {
 throw new Exception("Cannot pop() an empty stack!");
 }
 }

 public boolean empty()
 {
 return top == null;
 }

 }

}
</code></pre>

blocks|key|908761|text|public+static+void+main(String[]+args)+{+

++++String+str+="word";
++++if(str.equals(reverse(str)))+{
++++++++++System.out.println("Word+is+Palindromic+");
++++}else+{
++++++++++System.out.println("Word+is+Not+Palindromic+");
++++}
}


public+static+String+reverse(String+str)+{
++++char[]+charArr+=+str.toCharArray();
++++int+size+=+charArr.length;
++++Stack+stack+=+new+Stack(size);

++++int+i;
++++for(i+=+0;+i+<+size;+%2B%2Bi)+{
++++++++stack.push(charArr[i]);
++++}

++++for(i+=+0;+i+<+size;+%2B%2Bi)+{
++++++++charArr[i]+=+stack.pop();
++++}

++++return+String.valueOf(charArr);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|908762|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static void main(String[] args) { 

 String str ="word";
 if(str.equals(reverse(str))) {
 System.out.println("Word is Palindromic ");
 }else {
 System.out.println("Word is Not Palindromic ");
 }
}


public static String reverse(String str) {
 char[] charArr = str.toCharArray();
 int size = charArr.length;
 Stack stack = new Stack(size);

 int i;
 for(i = 0; i &lt; size; ++i) {
 stack.push(charArr[i]);
 }

 for(i = 0; i &lt; size; ++i) {
 charArr[i] = stack.pop();
 }

 return String.valueOf(charArr);
}
</code></pre>

I want to check a string is palindromic using stacks implement in java. I have tried many ways but its not getting me the correct answer. Can someone help me with it? 

I have implemented a Stack in my class which variable name is "stackArray". 

Palindromic method :

<pre><code>Stack s1 = new Stack(5);

for (int i = 0; i &lt; 5; i++) {
 s1.push(stackArray[i]);
}

String [] reverser = new String[5];

while (!s1.isEmpty()) {
 int count = 0;
 reverser[count] = s1.pop();
 count++;
}

if (s1.equals(reverser)){
 System.out.println("The input String is a palindrome.");
}
else
 System.out.println("The input String is not a palindrome.");
</code></pre>

Driver code :

<pre><code>public static void main(String[] args) throws Exception { 

 Stack st = new Stack(5);

 st.push("l");
 st.push("e");
 st.push("v");
 st.push("e");
 st.push("l");
 System.out.println("");
 System.out.println("When Palindromic is starting; ");
 System.out.println("");
 st.palindromic();
}
</code></pre>

Its obvious that the word "level" is Palindromic but always it says "The input String is not a palindrome."

Checking a string is palindromic using stacks

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想使用java中的堆栈实现来检查字符串是否回文。我尝试过许多方法，但这并没有给我正确的答案。有人能帮我吗？我在我的类中实现了一个变量名为"stackArray“的堆栈。回文法：Stack s1 = new Stack(5);for (int i = 0; i < 5; i++) { s1.push(stackA...

问使用堆栈检查字符串是回文。
EN

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问使用堆栈检查字符串是回文。EN