blocks|key|2530838|text|我怀疑有一个更有效的解决方案，但这里有一个天真的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2530839|将句子拆分成空格中的单词(注意-如果您有多个空格，我的实现将让problems)|2530840|Strip+punctuation|unordered-list-item|2530841|
|2530842|Reverse+|2530843|ordered-list-item|2530844|遍历每个字母，并从颠倒的单词中插入字符，并在必要时从原始单词中插入标点符号|2530845|2530846|2530847|public+class+Reverser+{

++++public+String+reverseSentence(String+sentence)+{
++++++++String[]+words+=+sentence.split("+");
++++++++return+Arrays.stream(words).map(this::reverseWord).collect(Collectors.joining("+"));
++++}

++++private+String+reverseWord(String+word)+{
++++++++String+noPunctuation+=+word.replaceAll("\\W",+"");
++++++++String+reversed+=+new+StringBuilder(noPunctuation).reverse().toString();
++++++++StringBuilder+result+=+new+StringBuilder();
++++++++for+(int+i+=+0;+i+<+word.length();+%2B%2Bi)+{
++++++++++++char+ch+=+word.charAt(i);
++++++++++++if+(!Character.isAlphabetic(ch)+&&+!Character.isDigit(ch))+{
++++++++++++++++result.append(ch);
++++++++++++}
++++++++++++if+(i+<+reversed.length())+{
++++++++++++++++result.append(reversed.charAt(i));
++++++++++++}
++++++++}
++++++++return+result.toString();
++++}
}|code-block|syntax|javascript|2530848|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|11|8|@]|9|@]|A|$]]|$1|K|3|-4|5|L|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|L|7|13|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|16|8|@]|9|@]|A|$T|U]]|$1|V|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|W|$]]

I suspect there's a more efficient solution but here's a naive one:

<ol>
<li>Split sentence into words on spaces (note - if you have multiple spaces my implementation will have problems)</li>
<li>Strip punctuation</li>
<li>Reverse each word</li>
<li>Go through each letter, and insert character from reversed word AND insert punctuation from original word if necessary</li>
</ol>

<pre><code>public class Reverser {

 public String reverseSentence(String sentence) {
 String[] words = sentence.split(" ");
 return Arrays.stream(words).map(this::reverseWord).collect(Collectors.joining(" "));
 }

 private String reverseWord(String word) {
 String noPunctuation = word.replaceAll("\\W", "");
 String reversed = new StringBuilder(noPunctuation).reverse().toString();
 StringBuilder result = new StringBuilder();
 for (int i = 0; i &lt; word.length(); ++i) {
 char ch = word.charAt(i);
 if (!Character.isAlphabetic(ch) &amp;&amp; !Character.isDigit(ch)) {
 result.append(ch);
 }
 if (i &lt; reversed.length()) {
 result.append(reversed.charAt(i));
 }
 }
 return result.toString();
 }
}
</code></pre>

blocks|key|2319518|text|不需要使用正则表达式，当然也不应该使用split("\\s%2B")，因为您会丢失连续的空格和空格字符的类型，即结果的空格可能是不正确的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2319519|您也不应该使用charAt()或任何类似的东西，因为这不支持来自Unicode+Supplemental平面的字母，即作为代理对存储在Java+string中的Unicode字符。|2319520|基本逻辑：|2319521|查找单词的开始，即字符串的开始或单词结束后的第一个字符，即空格之前的最后一个字符或从开始和结束平行开始的string.|2319522|Iterating的结束:忽略不是letters.|unordered-list-item|2319523|
|2319524|Swap+whitespace.|2319525|2319526|Locate+letters.|2319527|
的|2319528|2319529|2319530|2319531|2319532|​|2319533|作为Java代码，完全支持Unicode：|2319534|public+static+String+reverseLettersOfWords(String+input)+{
++++int[]+codePoints+=+input.codePoints().toArray();
++++for+(int+i+=+0,+start+=+0;+i+<=+codePoints.length;+i%2B%2B)+{
++++++++if+(i+==+codePoints.length+%7C%7C+Character.isWhitespace(codePoints[i]))+{
++++++++++++for+(int+end+=+i+-+1;+;+start%2B%2B,+end--)+{
++++++++++++++++while+(start+<+end+&&+!+Character.isLetter(codePoints[start]))
++++++++++++++++++++start%2B%2B;
++++++++++++++++while+(start+<+end+&&+!+Character.isLetter(codePoints[end]))
++++++++++++++++++++end--;
++++++++++++++++if+(start+>=+end)
++++++++++++++++++++break;
++++++++++++++++int+tmp+=+codePoints[start];
++++++++++++++++codePoints[start]+=+codePoints[end];
++++++++++++++++codePoints[end]+=+tmp;
++++++++++++}
++++++++++++start+=+i+%2B+1;
++++++++}
++++}
++++return+new+String(codePoints,+0,+codePoints.length);
}|code-block|syntax|javascript|2319535|测试|2319536|System.out.println(reverseLettersOfWords("Hello"));
System.out.println(reverseLettersOfWords("Hello!"));
System.out.println(reverseLettersOfWords("I'm+saying+Hello!"));
System.out.println(reverseLettersOfWords("I+haven't+said+hello+yet,+but+I+will."));
System.out.println(reverseLettersOfWords("Works+with+surrogate+pairs:+%2B+"));|2319537|输出|2319538|olleH
olleH!
m'I+gniyas+olleH!
I+tneva'h+dias+olleh+tey,+tub+I+lliw.
skroW+htiw+etagorrus+sriap:+%2B+|2319539|请注意，末尾的特殊字母是在“脚本(或书法)”列中显示的前4个字符，“here”，例如the+is+Unicode+Character+'MATHEMATICAL+BOLD+SCRIPT+CAPITAL+A'+(U%2B1D4D0)，在Java语言中是两个字符"\uD835\uDCD0"。|2319540|2319541|更新|BOLD|2319542|上面的实现是为颠倒单词的字母而优化的。要应用任意操作来损坏单词的字母，请使用以下实现：|2319543|public+static+String+mangleLettersOfWords(String+input)+{
++++int[]+codePoints+=+input.codePoints().toArray();
++++for+(int+i+=+0,+start+=+0;+i+<=+codePoints.length;+i%2B%2B)+{
++++++++if+(i+==+codePoints.length+%7C%7C+Character.isWhitespace(codePoints[i]))+{
++++++++++++int+wordCodePointLen+=+0;
++++++++++++for+(int+j+=+start;+j+<+i;+j%2B%2B)
++++++++++++++++if+(Character.isLetter(codePoints[j]))
++++++++++++++++++++wordCodePointLen%2B%2B;
++++++++++++if+(wordCodePointLen+!=+0)+{
++++++++++++++++int[]+wordCodePoints+=+new+int[wordCodePointLen];
++++++++++++++++for+(int+j+=+start,+k+=+0;+j+<+i;+j%2B%2B)
++++++++++++++++++++if+(Character.isLetter(codePoints[j]))
++++++++++++++++++++++++wordCodePoints[k%2B%2B]+=+codePoints[j];
++++++++++++++++int[]+mangledCodePoints+=+mangleWord(wordCodePoints.clone());
++++++++++++++++if+(mangledCodePoints.length+!=+wordCodePointLen)
++++++++++++++++++++throw+new+IllegalStateException("Mangled+word+is+wrong+length:+'"+%2B+new+String(wordCodePoints,+0,+wordCodePoints.length)+%2B+"'+("+%2B+wordCodePointLen+%2B+"+code+points)"+%2B
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++"+vs+mangled+'"+%2B+new+String(mangledCodePoints,+0,+mangledCodePoints.length)+%2B+"'+("+%2B+mangledCodePoints.length+%2B+"+code+points)");
++++++++++++++++for+(int+j+=+start,+k+=+0;+j+<+i;+j%2B%2B)
++++++++++++++++++++if+(Character.isLetter(codePoints[j]))
++++++++++++++++++++++++codePoints[j]+=+mangledCodePoints[k%2B%2B];
++++++++++++}
++++++++++++start+=+i+%2B+1;
++++++++}
++++}
++++return+new+String(codePoints,+0,+codePoints.length);
}
private+static+int[]+mangleWord(int[]+codePoints)+{
++++return+mangleWord(new+String(codePoints,+0,+codePoints.length)).codePoints().toArray();
}
private+static+CharSequence+mangleWord(String+word)+{
++++return+new+StringBuilder(word).reverse();
}|2319544|当然，如果需要，您可以将对任一mangleWord方法的硬编码调用替换为对传入的Function<int[],+int[]>或Function<String,+?+extends+CharSequence>参数的调用。|2319545|mangleWord方法实现的结果与原始实现相同，但是您现在可以轻松地实现不同的破坏算法。|2319546|例如，要随机化字母，只需对codePoints数组执行shuffle操作：|2319547|private+static+int[]+mangleWord(int[]+codePoints)+{
++++Random+rnd+=+new+Random();
++++for+(int+i+=+codePoints.length+-+1;+i+>+0;+i--)+{
++++++++int+j+=+rnd.nextInt(i+%2B+1);
++++++++int+tmp+=+codePoints[j];
++++++++codePoints[j]+=+codePoints[i];
++++++++codePoints[i]+=+tmp;
++++}
++++return+codePoints;
}|2319548|样本输出|2319549|Hlelo
Hlleo!
m'I+nsayig+oHlel!
I+athen'v+siad+eohll+yte,+btu+I+illw.
srWok+twih+rueoatrsg+rpasi:+%2B|2319550|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Mathematical_Alphanumeric_Symbols#Latin_letters|1|https://www.fileformat.info/info/unicode/char/1d4d0/index.htm|2|https://stackoverflow.com/a/1520212/5221149^0|J|D|0|7|8|0|0|0|0|0|0|0|0|0|1|1|0|0|0|0|0|0|0|0|0|3J|E|Y|4|0|1D|1S|1|0|0|0|2|0|0|0|F|A|14|M|1R|14|0|0|A|0|D|A|R|7|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|2G|8|@$9|2H|A|2I|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|2J|8|@$9|2K|A|2L|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|2M|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|2N|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|2O|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|2P|8|@]|D|@]|E|$]]|$1|Q|3|R|5|N|7|2Q|8|@]|D|@]|E|$]]|$1|S|3|P|5|6|7|2R|8|@]|D|@]|E|$]]|$1|T|3|U|5|N|7|2S|8|@]|D|@]|E|$]]|$1|V|3|W|5|6|7|2T|8|@]|D|@]|E|$]]|$1|X|3|-4|5|N|7|2U|8|@]|D|@]|E|$]]|$1|Y|3|-4|5|N|7|2V|8|@]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|2W|8|@]|D|@]|E|$]]|$1|10|3|-4|5|6|7|2X|8|@]|D|@]|E|$]]|$1|11|3|12|5|6|7|2Y|8|@]|D|@]|E|$]]|$1|13|3|14|5|6|7|2Z|8|@]|D|@]|E|$]]|$1|15|3|16|5|17|7|30|8|@]|D|@]|E|$18|19]]|$1|1A|3|1B|5|6|7|31|8|@]|D|@]|E|$]]|$1|1C|3|1D|5|17|7|32|8|@]|D|@]|E|$18|19]]|$1|1E|3|1F|5|6|7|33|8|@]|D|@]|E|$]]|$1|1G|3|1H|5|17|7|34|8|@]|D|@]|E|$18|19]]|$1|1I|3|1J|5|6|7|35|8|@$9|36|A|37|B|C]]|D|@$9|38|A|39|1|3A]|$9|3B|A|3C|1|3D]]|E|$]]|$1|1K|3|-4|5|6|7|3E|8|@]|D|@]|E|$]]|$1|1L|3|1M|5|6|7|3F|8|@$9|3G|A|3H|B|1N]]|D|@]|E|$]]|$1|1O|3|1P|5|6|7|3I|8|@]|D|@]|E|$]]|$1|1Q|3|1R|5|17|7|3J|8|@]|D|@]|E|$18|19]]|$1|1S|3|1T|5|6|7|3K|8|@$9|3L|A|3M|B|C]|$9|3N|A|3O|B|C]|$9|3P|A|3Q|B|C]]|D|@]|E|$]]|$1|1U|3|1V|5|6|7|3R|8|@$9|3S|A|3T|B|C]]|D|@]|E|$]]|$1|1W|3|1X|5|6|7|3U|8|@$9|3V|A|3W|B|C]]|D|@$9|3X|A|3Y|1|3Z]]|E|$]]|$1|1Y|3|1Z|5|17|7|40|8|@]|D|@]|E|$18|19]]|$1|20|3|21|5|6|7|41|8|@]|D|@]|E|$]]|$1|22|3|23|5|17|7|42|8|@]|D|@]|E|$18|19]]|$1|24|3|-4|5|6|7|43|8|@]|D|@]|E|$]]]|25|$26|$5|27|28|29|E|$2A|2B]]|2C|$5|27|28|29|E|$2A|2D]]|2E|$5|27|28|29|E|$2A|2F]]]]

No need to use regex for this, and you certainly shouldn't use <code>split("\\s+")</code>, since you'd lose consecutive spaces, and the type of whitespace characters, i.e. the spaces of the result could be incorrect.

You also shouldn't use <code>charAt()</code> or anything like it, since that would not support letters from the Unicode Supplemental Planes, i.e. Unicode characters that are stored in Java strings as surrogate pairs.

Basic logic:

<ul>
<li>Locate start of word, i.e. start of string or first character following whitespace.</li>
<li>Locate end of word, i.e. last character preceding whitespace or end of string.</li>
<li>Iterating from beginning and end in parallel:

<ul>
<li>Skip characters that are not letters.</li>
<li>Swap the letters.</li>
</ul></li>
</ul>

As Java code, with full Unicode support:

<pre><code>public static String reverseLettersOfWords(String input) {
 int[] codePoints = input.codePoints().toArray();
 for (int i = 0, start = 0; i &lt;= codePoints.length; i++) {
 if (i == codePoints.length || Character.isWhitespace(codePoints[i])) {
 for (int end = i - 1; ; start++, end--) {
 while (start &lt; end &amp;&amp; ! Character.isLetter(codePoints[start]))
 start++;
 while (start &lt; end &amp;&amp; ! Character.isLetter(codePoints[end]))
 end--;
 if (start &gt;= end)
 break;
 int tmp = codePoints[start];
 codePoints[start] = codePoints[end];
 codePoints[end] = tmp;
 }
 start = i + 1;
 }
 }
 return new String(codePoints, 0, codePoints.length);
}
</code></pre>

Test

<pre><code>System.out.println(reverseLettersOfWords("Hello"));
System.out.println(reverseLettersOfWords("Hello!"));
System.out.println(reverseLettersOfWords("I'm saying Hello!"));
System.out.println(reverseLettersOfWords("I haven't said hello yet, but I will."));
System.out.println(reverseLettersOfWords("Works with surrogate pairs: + "));
</code></pre>

Output

<pre class="lang-none prettyprint-override"><code>olleH
olleH!
m'I gniyas olleH!
I tneva'h dias olleh tey, tub I lliw.
skroW htiw etagorrus sriap: + 
</code></pre>

Note that the special letters at the end are the first 4 shown <a href="https://en.wikipedia.org/wiki/Mathematical_Alphanumeric_Symbols#Latin_letters" rel="nofollow noreferrer">here</a> in column "Script (or Calligraphy)", "Bold", e.g. the is <a href="https://www.fileformat.info/info/unicode/char/1d4d0/index.htm" rel="nofollow noreferrer">Unicode Character 'MATHEMATICAL BOLD SCRIPT CAPITAL A' (U+1D4D0)</a>, which in Java is two characters <code>"\uD835\uDCD0"</code>.

<hr>

UPDATE

The above implementation is optimized for reversing the letters of the word. To apply an arbitrary operation to mangle the letters of the word, use the following implementation:

<pre><code>public static String mangleLettersOfWords(String input) {
 int[] codePoints = input.codePoints().toArray();
 for (int i = 0, start = 0; i &lt;= codePoints.length; i++) {
 if (i == codePoints.length || Character.isWhitespace(codePoints[i])) {
 int wordCodePointLen = 0;
 for (int j = start; j &lt; i; j++)
 if (Character.isLetter(codePoints[j]))
 wordCodePointLen++;
 if (wordCodePointLen != 0) {
 int[] wordCodePoints = new int[wordCodePointLen];
 for (int j = start, k = 0; j &lt; i; j++)
 if (Character.isLetter(codePoints[j]))
 wordCodePoints[k++] = codePoints[j];
 int[] mangledCodePoints = mangleWord(wordCodePoints.clone());
 if (mangledCodePoints.length != wordCodePointLen)
 throw new IllegalStateException("Mangled word is wrong length: '" + new String(wordCodePoints, 0, wordCodePoints.length) + "' (" + wordCodePointLen + " code points)" +
 " vs mangled '" + new String(mangledCodePoints, 0, mangledCodePoints.length) + "' (" + mangledCodePoints.length + " code points)");
 for (int j = start, k = 0; j &lt; i; j++)
 if (Character.isLetter(codePoints[j]))
 codePoints[j] = mangledCodePoints[k++];
 }
 start = i + 1;
 }
 }
 return new String(codePoints, 0, codePoints.length);
}
private static int[] mangleWord(int[] codePoints) {
 return mangleWord(new String(codePoints, 0, codePoints.length)).codePoints().toArray();
}
private static CharSequence mangleWord(String word) {
 return new StringBuilder(word).reverse();
}
</code></pre>

You can of course replace the hardcoded call to the either <code>mangleWord</code> method with a call to a passed-in <code>Function&lt;int[], int[]&gt;</code> or <code>Function&lt;String, ? extends CharSequence&gt;</code> parameter, if needed.

The result with that implementation of the <code>mangleWord</code> method(s) is the same as the original implementation, but you can now easily implement a different mangling algorithm.

E.g. to randomize the letters, simply <a href="https://stackoverflow.com/a/1520212/5221149">shuffle</a> the <code>codePoints</code> array:

<pre><code>private static int[] mangleWord(int[] codePoints) {
 Random rnd = new Random();
 for (int i = codePoints.length - 1; i &gt; 0; i--) {
 int j = rnd.nextInt(i + 1);
 int tmp = codePoints[j];
 codePoints[j] = codePoints[i];
 codePoints[i] = tmp;
 }
 return codePoints;
}
</code></pre>

Sample Output

<pre class="lang-none prettyprint-override"><code>Hlelo
Hlleo!
m'I nsayig oHlel!
I athen'v siad eohll yte, btu I illw.
srWok twih rueoatrsg rpasi: +
</code></pre>

For instance, take the following list of <code>String</code>s, disregarding the inverted commas:

<pre><code>"Hello"
"Hello!"
"I'm saying Hello!"
"I haven't said hello yet, but I will."
</code></pre>

Now let's say I'd like to perform a certain operation on the characters of each word — for instance, say I'd like to reverse the characters, but keep the positions of the punctuation. So the result would be:

<pre><code>"olleH"
"olleH!"
"m'I gniyas olleH!"
"I tneva'h dias olleh tey, tub I lliw."
</code></pre>

Ideally I'd like my code to be independent of the operation performed on the string (another example would be a random shuffling of letters), and independent of all punctuation—so hyphens, apostrophes, commas, full stops, en/em dashes, etc. all remain in their original positions after the operation is performed. This probably requires some form of regular expressions. 

For this, I was thinking that I should save the indices and characters of all punctuation in a given word, perform the operation, and then re-insert all punctuation at their correct positions. However, I can't think of a way to do this, or a class to use.

I have a first attempt, but this unfortunately does not work with punctuation, which is the key: 

<pre><code>jshell&gt; String str = "I haven't said hello yet, but I will."
str ==&gt; "I haven't said hello yet, but I will."

jshell&gt; Arrays.stream(str.split("\\s+")).map(x -&gt; (new StringBuilder(x)).reverse().toString()).reduce((x, y) -&gt; x + " " + y).get()
$2 ==&gt; "I t'nevah dias olleh ,tey tub I .lliw"
</code></pre>

Has anyone got an idea how I might fix this? Thanks very much. There's no need for full working code—maybe just a signpost to an appropriate class I could use to perform this operation.

Modify the characters of words in a Java string with punctuation, but keep the positions of said punctuation?

Java

例如，以下面的String列表为例，忽略反逗号："Hello""Hello!""I'm saying Hello!""I haven't said hello yet, but I will."现在假设我想对每个单词的字符执行特定的操作-例如，假设我想颠倒字符，但保留标点符号的位置。因此，结果将是："olleH""ol...

问用标点符号修改Java字符串中单词的字符，但保留标点符号的位置？
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