如何解析集合的字符串？

Question

我正在編寫一種方法，該方法應該采用格式為"s1:{1,2,3,4}"的輸入String並將其放入Set 。 我自己開發的集合類，如下所示：

public class Set<E> implements Iterable<E> {
    private static final int DEFAULT_CAPACITY = 20;
    private String name;
    private E[] theData;
    private int size = 0;
    private int capacity = 0;

    public Set(){
        capacity = DEFAULT_CAPACITY;
        theData = (E[]) new Object[capacity];
    }//end constructor

    public Set(String name){
        capacity = DEFAULT_CAPACITY;
        theData = (E[]) new Object[capacity];
        this.name = name;
    }//end constructor

    public String getName(){
        return name;
    }//end getName

    public void setName(String name){
        this.name = name;
    }//end setName

    //adds object to set
    public void add(Object E) {
        if (size == capacity) {
            reallocate();
        }//end if

        theData[size] = (E) E;
        size++;

        for (int j = 0; j<size; j++) {
            for (int k = 0; k < size; k++) {
                if ((int)theData[j] < (int)theData[k]) {
                    E temp = theData[j];
                    theData[j] = theData[k];
                    theData[k] = temp; 
                }//end if
            }//end nested for loop
        }//end for loop

        int counter = 0;
        for (int i = 0; i < size; i++) {
            if (E == theData[i]) {
                counter++;

                if (counter >= 2) {
                    remove((Object)E);
                 }//end nested if
            }//end if
        }//end for loop
    }//end add method

    public E get(int i) {
        if (i < 0 || i >= size) {
            throw new ArrayIndexOutOfBoundsException(i);
        } else {
            return theData[i];
        }//end else 
    }//end get method

    public E remove(int i) {
        if (i < 0 || i >= size) {
            throw new ArrayIndexOutOfBoundsException(i);
        }//end if

        E returnValue = theData[i];
        for (int j = i + 1; j < size; j++) {
            theData[j - 1] = theData[j];
        }//end for loop

        size--;
        return returnValue;
    }//end remove method

    public void remove(Object E) {
        for (int i = 0; i < size; i++) {
            if (E == theData[i]) {
                for (int j = i + 1; j < size; j++){
                    theData[j - 1] = theData[j];
                }//end nested for loop

                size--;
            }//end if
        }//end for loop
    }//end remove method

    //fix!
    public int find(Object E) {
        int first, last, middle;
        first = 0;
        last = size - 1;
        middle = (first+last) / 2;

        while(first <= last ) {
            if ((int)theData[middle] > (int)E ) {
                last = middle - 1;    
            } else if ((int)theData[middle] < (int)E ) {
                first = middle + 1;
            } else {
                return middle;
            }//end else
        }//end while

        if (first > last) {
            return -1;
        }//end if

        return -1;
    }//end find method

    public Set<E> union(Set<E> s) {
        Set<E> returnSet = new Set<E>();

        for (int i = 0; i < this.size; i++) {
            returnSet.add(this.theData[i]);
        }//end for loop

        for (int i = 0; i < s.size; i++) {
            returnSet.add(s.theData[i]);
        }//end for loop

        return returnSet;
    }//end union method

    public Set<E> intersect(Set<E> s) {
        Set<E> returnSet = new Set<E>();

        for (int i = 0; i < this.size; i++) {
            for (int j = 0; j < s.size; j++) {
                if (this.theData[i] == s.theData[j]){
                    returnSet.add(theData[i]);
                }//end if
            }//end nested for loop
        }//end for loop

        return returnSet;
    }//end intersect method

    public Set<E> subtract(Set<E> s) {
        Set<E> returnSet = new Set<E>();

        for (int i = 0; i < this.size; i++) {
            for (int j = 0; j < s.size; j++) {
                if (this.theData[i] == s.theData[j]) {
                    this.remove((Object)this.theData[i]);
                    s.remove((Object)s.theData[j]);
                }//end if
            }//end nested for loop
        }//end for loop

        for (int i = 0; i < this.size; i++) {
            returnSet.add(this.theData[i]);
        }//end for loop

        for (int i = 0; i < s.size; i++) {
            returnSet.add(s.theData[i]);
        }//end for loop

        return returnSet;
    }//end subtract method

    public boolean equals(Set<E> s) {
        boolean result = false;

        for (int i = 0; i < this.size; i++) {
            if (this.theData[i] == s.theData[i]) {
                result = true;
            }//end if

            if (this.theData[i] != s.theData[i]) {
                result = false;
                break;
            }//end if
        }//end for loop

        return result;
    }//end equals method


    private void reallocate() {
        capacity = 2*capacity;
        theData = Arrays.copyOf(theData,  capacity);
    }//end reallocate method

    public String toString() {
        StringBuilder set = new StringBuilder();
        set.append("{");

        for (int i = 0; i < size; i++) {
            set.append(theData[i]);

            if (i != size-1){
                set.append(",");
            }//end if
        }//end for loop

        set.append("}");

        return set.toString();
    }//end toString()

    public SetIterator<E> iterator() {
        SetIterator<E> it = new SetIterator<E>() {
            private int currentIndex = 0;

            public boolean hasNext() {
                if (currentIndex < size && theData[currentIndex] != null){
                    currentIndex++;
                    return true;
                } else{
                    return false;
                }//end else
            }//end hasNext()


            public E next() {
                if (!hasNext()) {
                    throw new NoSuchElementException();
                }//end if

                return theData[currentIndex++];
            }//end next()


            public boolean hasPrevious() {
                if (currentIndex <= size && currentIndex > 0) {
                    currentIndex--;
                    return true;
                } else {
                    return false;
                }//end else
            }//end hasPrevious()


            public E previous() {
                if (!hasPrevious()) {
                    throw new NoSuchElementException();
                }//end if

                return theData[currentIndex--];
            }//end previous()


            public void add(E item) {
                theData[currentIndex-1] = item;
            }//end add()


            public void remove() {
                for (int i = 0; i < size; i++) {
                    if (theData[currentIndex] == theData[i]) {
                        for (int j = i + 1; j < size; j++) {
                            theData[j - 1] = theData[j];
                        }//end nested for loop

                        size--;
                    }//end if
                }//end for loop
            }//end remove()
        };//end new SetIterator()

        return it;
    }//end iterator method
}//end Set class

方法是應該

• 如果該方法的格式無效，例如"s1:[1 2,3,4}" （ 此示例為缺少逗號和花括號 ），則引發異常。
• 此外，輸入可以具有任意數量的空格，並且仍然被視為有效。 示例： "s1: {1, 2, 3, 4 }" 。

到目前為止，我所擁有的方法是：

public Set<Integer> parse(String input){
    String s[] = input.split(":");
    String name = s[0];
    Set<Integer> returnSet = new Set<Integer>(name);

    return returnSet;
}

我不確定如何正確地從字符串集中檢索元素並將其放入Set對象。 我知道我自己獲得它們parseInt可以parseInt ，但是在隔離每個元素時遇到了麻煩。 一個集合可以有多少個元素沒有限制。 這意味着我的代碼應該可以使用任意數量的元素。

我也考慮過正則表達式，但是我覺得似乎有一種更有效的方法。

任何幫助將不勝感激！

Answer 1

我給了您最少的代碼。 這將匹配或返回null。 然后，您將獲得標簽和字符串集。 如果確實需要Integer對象，則可以像下面的f2（）一樣簡單地進行轉換。 您需要添加的是錯誤處理和更多注釋。 請查看JavaDoc API，以獲取有關Pattern / Matcher的更多信息。 另外，不要簡單地使用HashSet。 如果訂單對您很重要，則至少需要一個LinkedHashSet。 如果允許重復，請勿使用任何哈希值！ 使用LinkedList或數組。

順便說一句，您分割字符串的方法沒有錯，但是會更復雜。 您必須除以：，然后調用str.trim（）刪除任何多余的空格，str.substring（startIndx，endIndex），然后最后可以解析數字列表。 您將必須使用str.indexOf（“ {”）或手動搜索以獲取索引。

import java.util.Arrays;
import java.util.LinkedHashSet;
import java.util.Set;
import java.util.regex.Matcher;
import java.util.regex.Pattern;


public class NewClass {

    //match this
    //STR:{NUM_LIST}
    //[A-Za-z0-9_]+ = STR is upper and lower alpha, number or underscore; 1 or more characters (in any order)
    //[0-9,]+ = NUM_LIST is one or more characters and can only contain numbers or comma (in any order)
    //The () used will give us a group
    //I like to explicitly use [] to specify a character, but it may not be needed
    //use a slash (needs a 2nd because of Java) to make sure it is interpreted as just a character and not as a structure of syntax.
    Pattern p=Pattern.compile("([A-Za-z0-9_]+)[:][\\{]([0-9,]+)[\\}]");

    Set test(String txt){
        Matcher m=p.matcher(txt);
        if(!m.matches())return null;
        int groups=m.groupCount();//should only equal 3 (default whole match+2groups) here, but you can test this

        System.out.println("Matched: " + m.group(0));
        String label = m.group(1);
        String[] arr = m.group(2).split(",");

        Set<String> set = new LinkedHashSet(Arrays.asList(arr));

        return set;
    }

     Object[] test2(String txt){
        Matcher m=p.matcher(txt);
        if(!m.matches())return null;
        int groups=m.groupCount();//should only equal 3 (default whole match+2groups) here, but you can test this

        System.out.println("Matched: " + m.group(0));
        String label = m.group(1);
        String[] arr = m.group(2).split(",");

        Set<String> set = new LinkedHashSet(Arrays.asList(arr));
        Object[] ret=new Object[3];
        ret[0] = m.group(0);
        ret[1] = label;
        ret[2] = set;
        return ret;
    }
}


 void f2(String[] arr){
     ArrayList<Integer> list=new ArrayList<Integer>(1000);
     for(String s: arr){
         try {
             list.add(Integer.parseInt(s));
         } catch (NumberFormatException numberFormatException) {
             System.out.println(numberFormatException+ "\t-->\t"+ s);
         }
     }
    Set<Integer> set = new LinkedHashSet(list);
 }

Answer 2

最簡單的方法是使用Set的構造函數

http://docs.oracle.com/javase/7/docs/api/java/util/HashSet.html

連同Arrays.asList()

http://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html

將您的String[]轉換為Set<String> ：

Set<String> mySet = new HashSet<String>(Arrays.asList(s));

Answer 3

這是一個有效的示例：首先，您將創建一個正則表達式模式以匹配{}的內部，然后檢查{}的內部，如果其格式正確。 然后，您將{}的內部轉換為ArrayList。

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.*;

public class Test {

    public String test = "s 1 : {1, 2,3 ,4}";

    public Test() {
        //match the inside of {}
        Pattern pattern = Pattern.compile("^s\\s*\\d+\\s*:\\s*\\{([0-9,\\s*]*)}");

        Matcher matcher = pattern.matcher(test);


        // check all occurance
        while (matcher.find()) {
            if(matcher.group(1).trim().matches("^(\\d*)+(\\s*,\\s*\\d*)*$")) {
                System.out.println("valid string");
                List<String> items = Arrays.asList(matcher.group(1).split("\\s*,\\s*"));
                for(String number: items) {
                    System.out.println(number.trim());
                }
            }else{
                System.out.println("invalid string");
            }
        }
    }

    public static void main(String[] args) {
        new Test();
    }
}