有没有办法替换列表变量中所有出现的字符串？

Question

I have a list as follows

List<Summary> summary;

Summary {
    int key;
    String value1;
    String value2;
}

Is there a way in java (or any other utility library) to replace all occurrences of a String "String1" in the variable value1 without having to loop through the list?

Answer 1

Looping the list is inevitable so you or some third party library has to do the looping. Is it really that hard to do:

for (Summary s : summary)
    if (s.value1.equals("String1"))
        s.value1 = "...";

? :)

Answer 2

Maybe you could use a library that allow you use it without loops, the problem is that in the low level the compiler must use something like a loop for do it. I think that direct or indirectly you will use a loop.

So, for this reason you haven´t any problem if use a loop in your code.

Sorry for my English, I hope you can understand all.

Answer 3

You can add a method to find by the object part:

public class ListOfString {

    public static void main(String[] args) {
        List<Model> models = new ArrayList<>();

        for(int i = 0 ; i < 5; i++) {
            Model model = new Model();
            model.setStr("String"+i);
            models.add(model);
        }

        Model temp = new Model();
        temp.setStr("String1");

        System.out.println(containsObjectPart(temp, models));
    }

    // This method just a prototype, you can modify as you like...
    private static boolean containsObjectPart(Model obj, List<Model> models) {
        for(Model model : models) {
            if(model.getStr().equals(obj.getStr()))
                return true;
        }

        return false;
    }

}

class Model {

    private String str;

    public String getStr() {
        return str;
    }

    public void setStr(String str) {
        this.str = str;
    }
}

Answer 4

不，你不能， Loop是invetible 。但是你可以在很多方面优化它，例如selecting loop ， how do you find the string and replacing it

Answer 5

You could avoid a bit of looping - at the expense of losing your list ordering - by using a TreeSet , with value1 values as keys and the Summary objects as values. Then you can do a binary chop search to find all matching entries. A collection which uses hashing would have similar trade-offs and gains.

Otherwise, as everyone else has said, looping is inevitable. The only optimisation you could make is counting the entries as they go into the list so you know when you've found them all so you can stop looping.

But remember, premature optimisation is the root of all evil . :)