Question

我正在开发一个分析文本的短程序，但是我必须从Arraylist中删除所有使用过的字母的所有spezial字母。

我该怎么做？我的三：

String[] SpezialLetters = {"1"," ","2","3","4","5","6","7","8","9","0","^","°","!","§","$","%","&","/","{","(","[",")","]","=","}","?","´","`","+","*","~","#","'","<",">","|","@","€",",",";",".",":","-","_"};
List<String> Letters = new ArrayList<String>(Arrays.asList(textToAnalyse.split("")));

    for(String Letter : Letters){
        if (Arrays.asList(SpezialLetters) != null && Arrays.asList(SpezialLetters).contains(Letter)){
           Letters.remove(Letter);
        }
    }

Answer 1

letters.removeAll(Arrays.asList(spezialLetters));应该这样做。

Answer 2

试试这个：

for (int i = 0; i < letters.size(); i++) {
    if (letters.get(i).replaceAll("[^a-zA-Z0-9]+","").equals("")){
       letters.remove(i);
    }
}

Answer 3

而不是在List中创建文字的Strings个字符textToAnalyse。以下代码段不是更好的解决方案吗？

String toRemove = "[ \\d\\^°!§$%&/{}()\\[\\]=\\?´`\\+\\*~#'<>\\|@€,;\\.:\\-_]";
String textToAnalyse = "some f00 un _-:.;,€@|><'#~*+`´?}=])[({/&%$§!°^0987654321 ";
String cleaned = textToAnalyse.replaceAll(toRemove, "");
System.out.printf("toRemove     : [%s]%n", toRemove);
System.out.printf("textToAnalyse: [%s]%n", textToAnalyse);
System.out.printf("cleaned      : [%s]%n", cleaned);

<强>输出

toRemove     : [ \d\^°!§$%&/{}()\[\]=\?´`\+\*~#'<>\|@€,;\.:\-_]
textToAnalyse: [some f00 un _-:.;,€@|><'#~*+`´?}=])[({/&%$§!°^0987654321 ]
cleaned      : [somefun]

toRemove模式的含义：

\d - 匹配所有数字
^[]?+*|.- - 这些字符在正则表达式pattern中具有特殊含义，因此必须转义
all others - 匹配文字字符

Java - 如何从Arraylist中删除Array的所有元素

3 个答案: