blocks|key|1865179|text|当您调用word.reverse()时，它不会返回反向StringBuilder的新实例。相反，它是在StringBuilder中反转元素，因此word和reverseWord+StringBuilder对象都有相反顺序的字符。您不需要使用两个StringBuilder对象。您只需要一个来找出相反的情况，然后您就可以将它与原语进行如下比较：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1865180|String+input+=+scanner.nextLine();
StringBuilder+reverseWordBuilder+=+new+StringBuilder(input);

String+reversedWord+=+reverseWordBuilder.reverse().toString();
if(input.equals(reversedWord))+{
++++System.out.println("yes");
}+else+{
++++System.out.println("no");
}|code-block|syntax|javascript|1865181|entityMap^0|4|E|R|D|1F|D|20|4|25|B|2H|D|3E|D|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|L|$]]

When you're calling <code>word.reverse()</code> it's not returning a new instance of reversed <code>StringBuilder</code>. Instead it's reversing elements inside <code>StringBuilder</code> so both <code>word</code> and <code>reverseWord</code> <code>StringBuilder</code> objects have characters in reversed order. You don't need to use two <code>StringBuilder</code> objects. You just need one to find out the reverse and then you can compare it with orignal like this:
<pre><code>String input = scanner.nextLine();
StringBuilder reverseWordBuilder = new StringBuilder(input);

String reversedWord = reverseWordBuilder.reverse().toString();
if(input.equals(reversedWord)) {
 System.out.println(&quot;yes&quot;);
} else {
 System.out.println(&quot;no&quot;);
}
</code></pre>

blocks|key|2706037|text|对word.reverse()的调用返回this，而不是StringBuilder的新实例-因此您将看到word+==+reverseWord。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2706038|如果将System.out.println(word);移动到word.reverse()调用下，则可以确认这一点，并且两行都将打印相同(反向)值。|2706039|entityMap^0|1|E|K|4|1G|J|0|3|P|V|E|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$]]

The call to <code>word.reverse()</code> returns <code>this</code> and not a new instance of StringBuilder - so you'll see that <code>word == reverseWord</code>.
You can confirm this if you move the <code>System.out.println(word);</code> under the call to <code>word.reverse()</code> and both lines will print the same (reversed) value.

blocks|key|1865205|text|StringBuilder是可变的，这意味着您可以更改他的状态，而无需创建它的新实例。String是一个不变的，它意味着每当您做一些更改时，就会在heap中创建一个具有不同地址的新对象。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|1865206|entityMap^0|0|D|17|6|21|4|E|3|1G|3|0^^$0|@$1|2|3|4|5|6|7|I|8|@$9|J|A|K|B|C]|$9|L|A|M|B|C]|$9|N|A|O|B|C]|$9|P|A|Q|B|D]|$9|R|A|S|B|D]]|E|@]|F|$]]|$1|G|3|-4|5|6|7|T|8|@]|E|@]|F|$]]]|H|$]]

<code>StringBuilder</code> is mutable, that means you can change his state without creating a new instance of it. <code>String</code> is a immutable that means every time when you makes some change, it is created a new object whit a different address in <code>heap</code>.

blocks|key|1316241|text|构建器|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1316242|StringBuilder严格遵循构建器模式。|offset|length|style|CODE|1316243|构建器模式方法主要用于返回对同一个对象的引用(如return+this)。|1316244|在您的示例中，word.reverse()将返回对同一个对象的引用。|1316245|虽然有两个不同的引用变量，但它们都指向同一个对象，因此任何类似的比较都将返回true。|1316246|更清楚的是，word+==+reverseWord也将是true|1316247|entityMap^0|0|0|D|0|O|B|0|7|E|0|12|4|0|6|J|S|4|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|14|8|@$D|15|E|16|F|G]|$D|17|E|18|F|G]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|Q|$]]

<h2>Builder</h2>
<code>StringBuilder</code> closely follows the builder pattern.
Builder pattern methods are primarily implemented to return the reference to same object(like <code>return this</code>).
In your example, <code>word.reverse()</code> will return the reference to the same object.
Though there are 2 different reference variables, both of them are pointing to the same object and hence any similar comparison will return <code>true</code>
To be more clear, <code>word == reverseWord</code> will be <code>true</code> too

blocks|key|1316309|text|当一个新的Heap+Memory.|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1316310|During对象被创建时，它被放置在对象的一个操作中，没有像经典的String那样创建新的对象，所以在任何操作之后，实际的对象都会被修改。|unordered-list-item|1316311|1316312|解释回文：|1316313|在本例中，假设是|1316314|，假设是word+=+"AS0A"。在Heap中，假设它指向@00AAXX对象.|ordered-list-item|1316315|1316316|1316317|当您调用word.reverse()时，它会反转"AS0A“>+"A0SA”。并在word中更新Heap对象。现在，Heap.|1316318|中的word+=+"A0SA"|1316319|1316320|现在，当您将word.reverse()分配给StringBuilder+reverseWord时，这次不会创建新的对象。reverseWord只要指向@00AAXX(word对象的位置).|1316321|1316322|如果您想要在堆中创建一个新对象，请使用|1316323|。那么您应该使用new：|1316324|1316325|StringBuilder+reverseWord+=+new+StringBuilder(word.reverse());|code-block|syntax|javascript|1316326|但是，|1316327|+new+object并没有解决这个问题，因为word对象的值已经被修改了。|1316328|1316329|1316330|因此，在调用反向之前，存储值不同的位置。然后比较.|1316331|1316332|修改后的代码块：|1316333|public+static+void+main(String[]+args)+{
++++++++Scanner+scanner+=+new+Scanner(System.in);

++++++++StringBuilder+word+=+new+StringBuilder(scanner.nextLine());
++++++++System.out.println(word);
++++++++String+wordString+=+word.toString();
++++++++StringBuilder+reverseWord+=+new+StringBuilder(word.reverse());
++++++++
++++++++System.out.println(reverseWord);
++++++++System.out.println(word);

++++++++if(wordString.equals(reverseWord.toString()))+{
++++++++++++System.out.println("yes");
++++++++}else+{
++++++++++++System.out.println("no");
++++++++}
}|1316334|entityMap^0|5|B|0|X|6|0|0|0|0|4|D|J|4|U|7|0|0|0|4|E|15|4|1C|4|1M|4|0|2|D|0|0|6|E|N|P|1P|B|24|7|2C|4|0|0|0|8|3|0|0|0|0|1|A|N|4|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1P|8|@$9|1Q|A|1R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1S|8|@$9|1T|A|1U|B|C]]|D|@]|E|$]]|$1|I|3|-4|5|6|7|1V|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|1W|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1X|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|1Y|8|@$9|1Z|A|20|B|C]|$9|21|A|22|B|C]|$9|23|A|24|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|25|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|26|8|@]|D|@]|E|$]]|$1|S|3|T|5|P|7|27|8|@$9|28|A|29|B|C]|$9|2A|A|2B|B|C]|$9|2C|A|2D|B|C]|$9|2E|A|2F|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|2G|8|@$9|2H|A|2I|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|2J|8|@]|D|@]|E|$]]|$1|X|3|Y|5|P|7|2K|8|@$9|2L|A|2M|B|C]|$9|2N|A|2O|B|C]|$9|2P|A|2Q|B|C]|$9|2R|A|2S|B|C]|$9|2T|A|2U|B|C]]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|2V|8|@]|D|@]|E|$]]|$1|10|3|11|5|6|7|2W|8|@]|D|@]|E|$]]|$1|12|3|13|5|P|7|2X|8|@$9|2Y|A|2Z|B|C]]|D|@]|E|$]]|$1|14|3|-4|5|6|7|30|8|@]|D|@]|E|$]]|$1|15|3|16|5|17|7|31|8|@]|D|@]|E|$18|19]]|$1|1A|3|1B|5|6|7|32|8|@]|D|@]|E|$]]|$1|1C|3|1D|5|P|7|33|8|@$9|34|A|35|B|C]|$9|36|A|37|B|C]]|D|@]|E|$]]|$1|1E|3|-4|5|6|7|38|8|@]|D|@]|E|$]]|$1|1F|3|-4|5|6|7|39|8|@]|D|@]|E|$]]|$1|1G|3|1H|5|P|7|3A|8|@]|D|@]|E|$]]|$1|1I|3|-4|5|6|7|3B|8|@]|D|@]|E|$]]|$1|1J|3|1K|5|6|7|3C|8|@]|D|@]|E|$]]|$1|1L|3|1M|5|17|7|3D|8|@]|D|@]|E|$18|19]]|$1|1N|3|-4|5|6|7|3E|8|@]|D|@]|E|$]]]|1O|$]]

<ul>
<li>When a new <code>StringBuilder</code> object is created, it is placed in <code>Heap Memory</code>.</li>
<li>During an operation on the <code>StringBuilder</code> object, no new object is created like classical <code>String</code>, so after any operation, the actual object is modified.</li>
</ul>
Explain palindrome senario:
<ol>
<li>In this case, suppose <code>word = &quot;AS0A&quot;</code>. In <code>Heap</code>, suppose it points <code>@00AAXX</code> object.
</li>
<li>When you call <code>word.reverse()</code>, it reverses &quot;AS0A&quot; &gt; &quot;A0SA&quot;. And update <code>word</code> object in the <code>Heap</code>. Now, <code>word = &quot;A0SA&quot;</code> in <code>Heap</code>.
</li>
<li>Now, when you assign <code>word.reverse()</code> into <code>StringBuilder reverseWord</code>, this time no new object will create. <code>reverseWord</code> just point <code>@00AAXX</code>(<code>word</code> object's location).
</li>
<li>If you want a new object to create in Heap. You should use <code>new</code> then:
</li>
</ol>
<pre><code>StringBuilder reverseWord = new StringBuilder(word.reverse());
</code></pre>
<ol start="5">
<li>But, <code>new object</code> does not solve this issue, as the value of <code>word</code> object already modified.
</li>
<li>So, before call reverse, store value different location. And then compare.
</li>
</ol>
Modified code block:
<pre><code>public static void main(String[] args) {
 Scanner scanner = new Scanner(System.in);

 StringBuilder word = new StringBuilder(scanner.nextLine());
 System.out.println(word);
 String wordString = word.toString();
 StringBuilder reverseWord = new StringBuilder(word.reverse());
 
 System.out.println(reverseWord);
 System.out.println(word);

 if(wordString.equals(reverseWord.toString())) {
 System.out.println(&quot;yes&quot;);
 }else {
 System.out.println(&quot;no&quot;);
 }
}
</code></pre>

I am trying to learn using StringBuilder class with a palindrome example and written this:
<pre><code>public static void main(String[] args) {
 Scanner scanner = new Scanner(System.in);

 StringBuilder word = new StringBuilder(scanner.nextLine());
 System.out.println(word);
 StringBuilder reverseWord = word.reverse();
 System.out.println(reverseWord);

 if(word.toString().equals(reverseWord.toString())) {
 System.out.println(&quot;yes&quot;);
 }else {
 System.out.println(&quot;no&quot;);
 }
}
</code></pre>
Program constantly gives me the result &quot;yes&quot; but I fail to understand why? Obviously there is something wrong on my if block but cannot see what it is.

Using StringBuilder for Palindrome Example in Java

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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我试图通过一个回文示例学习如何使用StringBuilder类，并编写如下：public static void main(String[] args) {    Scanner scanner = new Scanner(System.in);    StringBuilder word = new StringBuilder(scanner.nextLine());    System.out

问用StringBuilder实现回文实例
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