在Java中，有一个更优雅的方式来删除重复的字符串和ArrayList字符串的？ [英] In Java, is there a more elegant way to remove duplicate strings from and ArrayList of Strings?

问题描述

所以，长话短说，我有一个需要字符串长ArrayList的一个Java的家庭作业以各种方式来操纵（我们正在做这样的事情的话显示出组合，添加和ArrayList中，没有什么太移除特别）。我注意到，一些提供的ArrayList中有重复的条目（和重复是没有必要为这个任务），所以我就从我的老师还好通过删除重复项消毒的数据。
以下是我想出了：

So, a long story short, I have a Java homework assignment that requires a long ArrayList of Strings to be manipulated in various ways (we're doing things like showing combinations of words, adding and removing from the ArrayList, nothing too special). I noticed that a few of the provided ArrayLists have duplicate entries (and the duplicates aren't necessary for this assignment), so I got the okay from my teacher to sanitize the data by removing duplicate entries. Here's what I came up with:

private static ArrayList<String> KillDups(ArrayList<String> ListOfStrings) {  

    for (int i = 0 ; i < ListOfStrings.size(); i++) {
        for (int j = i + 1; j < ListOfStrings.size(); j++) {
            //don't start on the same word or you'll eliminate it.
            if ( ListOfStrings.get(i).toString().equalsIgnoreCase( ListOfStrings.get(j).toString() )  ) {
                ListOfStrings.remove(j);//if they are the same, DITCH ONE.
                j = j -1; //removing the word basically changes the index, so swing down one.
            }                                
        }
    }
    return ListOfStrings;
}

这是罚款，我的任务，但我怀疑它会在现实世界中非常有用的。有没有办法做到这一点，将忽略比较期间的空白和特殊字符？是否有一个普遍更清洁的方式（不嵌套For循环也许）来处理呢？难道还有其他的问题，我应该问，我不知道要问？

This is fine for my assignment, but I doubt it would be very useful in the real world. Is there a way to do this that would ignore white space and special characters during the comparison? Is there a cleaner way in general to handle this (maybe without the nested For Loops)? Is there another question I should be asking that I don't know to ask?

推荐答案

是的。它可以在短短1（高雅）行完成：

Yes. And it can be done in just 1 (elegant) line:

List<String> noDups = new ArrayList<String>(new LinkedHashSet<String>(list));

中间 设置 保证没有重复。该 LinkedHashSet 实施设置的被选为preserve列表的顺序。

The intermediate Set ensures no duplicates. The LinkedHashSet implementation of Set was chosen to preserve the order of the list.

此外，在一个样式注意：

Also, on a style note:

• 您的名字方法和参数的开头以小写字母名称


• 总是指抽象的（即列表），而不是具体的（即的ArrayList ）指定的方法签名时，

• name your methods and parameters with names starting with a lowercase letter
• always refer to the abstract (ie List) rather than the concrete (ie ArrayList) when specifying method signatures

您整个的方法则是：

private static List<String> killDups(List<String> list) {
    return new ArrayList<String>(new LinkedHashSet<String>(list));
}

有关额外印象分使该方法通用的，所以它的任何作品的类型列表：

For extra brownie points make the method generic, so it works with any type of List:

private static <T> List<T> killDups(List<T> list) {
    return new ArrayList<T>(new LinkedHashSet<T>(list));
}

如果你想忽略某些字符，我想创建一个类，并将这些名单。无论是 散code（） 和的 等于（） 方法经的 HashSets 来删除的DUP：

If you wanted to ignore certain characters, I'd create a class for that and have a list of those. Both the hashCode() and the equals() methods are relied upon by HashSets to remove dups:

public class MungedString {
    // simplified code
    String s;

    public boolean equals(Object o) {
        // implement how you want to compare them here
    }

    public int hashCode() {
        // keep this consistent with equals()
    }
}

然后

List<MungedString> list;
List<MungedString> noDupList = killDups(list);

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