英文:
Java: How to reverse string based on non alphabetic delimiters?
问题
String fileContent = "Joe'); MAKE TEST random;--";
String[] splitWords = fileContent.split("[^a-zA-Z0-9']+");
StringBuilder stringBuilder = new StringBuilder();
for (String word : splitWords) {
int idx = fileContent.indexOf(word, stringBuilder.length());
String delim = fileContent.substring(stringBuilder.length(), idx);
stringBuilder.append(delim);
StringBuilder output = new StringBuilder(word).reverse();
stringBuilder.append(output);
}
return stringBuilder.toString();
Current output: 'eoJ); EKAM TSET modnar
Desired output: eoJ'); EKAM TSET modnar;--
<details>
<summary>英文:</summary>
How to reverse string based on non alphabetic delimiters? I suspect my regex may be the problem.
String fileContent = "Joe'); MAKE TEST random;--";
String[] splitWords = fileContent.split("[^a-zA-Z0-9']+");
StringBuilder stringBuilder = new StringBuilder();
for (String word : splitWords) {
int idx = fileContent.indexOf(word, stringBuilder.length());
String delim = fileContent.substring(stringBuilder.length(), idx);
stringBuilder.append(delim);
StringBuilder output = new StringBuilder(word).reverse();
stringBuilder.append(output);
}
return stringBuilder.toString();
Current output: `'eoJ); EKAM TSET modnar`
Desired output: `eoJ'); EKAM TSET modnar;--`
</details>
# 答案1
**得分**: 2
你不需要正则表达式来实现这个。看起来你想要翻转仅为字母或数字的字符。然后你可以按照以下方式操作 - 获取字符数组中找到的字符是字母或数字的起始和结束索引，然后原地翻转。然后返回一个新的字符串，其中字符被原地翻转。
```java
private static void reverseWords(char[] c) {
int start = 0, end = c.length;
while (start < end) {
int pre = start;
while (start < c.length && Character.isLetterOrDigit(c[start]))
start++;
if (pre < start)
reverseWord(c, pre, start - 1);
start++;
}
}
private static void reverseWord(char[] c, int start, int end) {
while (start < end) {
char temp = c[start];
c[start] = c[end];
c[end] = temp;
start++;
end--;
}
}
你可以在这里测试这段代码。
英文:
You don't need a regex for this. It seems you want to reverse characters that are only alphabets or digits.
Then you could do this way - get the start and end indices of the character array where you find the character to be a letter or digit and then reverse in place. Then return a new String with the characters reversed in place.
private static void reverseWords(char[] c) {
int start = 0, end = c.length;
while ( start < end ) {
int pre = start;
while ( start < c.length && Character.isLetterOrDigit(c[start]) )
start++;
if ( pre < start )
reverseWord(c, pre, start-1);
start++;
}
}
private static void reverseWord(char[] c, int start, int end) {
while ( start < end ) {
char temp = c[start];
c[start] = c[end];
c[end] = temp;
start++;
end--;
}
}
You can test this code here
答案2
得分: 1
以下是翻译好的内容：
您可以匹配和反转仅包含1个或更多字母的块（使用简单的`\p{L}+`模式），并保留其余部分不变：
String s = "Joe'); MAKE TEST random;--";
StringBuffer result = new StringBuffer();
Matcher m = Pattern.compile("\\p{L}+").matcher(s);
while (m.find()) {
String replacement = new StringBuilder(m.group()).reverse().toString();
m.appendReplacement(result, replacement);
}
m.appendTail(result);
System.out.println(result.toString()); // => eoJ'); EKAM TSET modnar;--
请参阅在线Java演示。
英文:
You may match and reverse only chunks of 1+ letters (with a simple \p{L}+ pattern) and keep the rest as is:
String s = "Joe'); MAKE TEST random;--";
StringBuffer result = new StringBuffer();
Matcher m = Pattern.compile("\\p{L}+").matcher(s);
while (m.find()) {
String replacement = new StringBuilder(m.group()).reverse().toString();
m.appendReplacement(result, replacement);
}
m.appendTail(result);
System.out.println(result.toString()); // => eoJ'); EKAM TSET modnar;--
See the Java demo online.
答案3
得分: 1
你的代码在两处做了修改：
- 替换模式
- 检测尾部分隔符
String fileContent = "Joe'); MAKE TEST random;--";
String[] splitWords = fileContent.split("\\W"); // W是非单词字符，或者我忘了
StringBuilder stringBuilder = new StringBuilder();
for (String word : splitWords) {
int idx = fileContent.indexOf(word, stringBuilder.length());
String delim = fileContent.substring(stringBuilder.length(), idx);
stringBuilder.append(delim);
StringBuilder output = new StringBuilder(word).reverse();
stringBuilder.append(output);
}
// 是否有尾部分隔符？
if (fileContent.length() != stringBuilder.length()) {
// 添加剩余部分
stringBuilder.append(fileContent.substring(stringBuilder.length()));
}
return stringBuilder.toString();
英文:
Your code works with two changes:
- Replace the pattern
- Test for trailing delimiter
String fileContent = "Joe'); MAKE TEST random;--";
String[] splitWords = fileContent.split("\W"); // W is non word character or
I forgot
StringBuilder stringBuilder = new StringBuilder();
for (String word : splitWords) {
int idx = fileContent.indexOf(word, stringBuilder.length());
String delim = fileContent.substring(stringBuilder.length(), idx);
stringBuilder.append(delim);
StringBuilder output = new StringBuilder(word).reverse();
stringBuilder.append(output);
}
// did we have trailing delimiter ?
if(fileContent.length()!=stringBuilder.length())
{ //append remaining
stringBuilder.append(fileContent.substring(stringBuilder.length()));
}
return stringBuilder.toString();
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