我正在编写一个方法，该方法应该采用格式String的输入"s1:{1,2,3,4}"并将其放入Set。我自己开发的集合类，如下：
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
方法应该
"s1:[1 2,3,4}"（此示例是缺少的逗号和大括号），则抛出异常。 "s1: {1, 2, 3, 4 }"。到目前为止，我所拥有的方法是：
public Set<Integer> parse(String input){
String s[] = input.split(":");
String name = s[0];
Set<Integer> returnSet = new Set<Integer>(name);
return returnSet;
}
我不确定如何正确地从字符串中的集合中检索元素并将它们放入Set对象中。我知道我可以parseInt一旦我自己得到它们但我无法隔离每个元素。一组可以有多少元素没有限制;这意味着我的代码应该可以使用任意数量的元素。
我也考虑过正则表达式，但我觉得有一种更有效的方法。
任何帮助将不胜感激！
答案 0 :(得分：1)
我已经为您提供了所需的最少代码。这将匹配或返回null。然后你得到标签和字符串集。如果你真的需要Integer对象，你可以像下面的f2（）一样简单转换。您需要添加的是错误处理和更多注释。有关Pattern / Matcher的更多信息，请查看JavaDoc API。另外，不要只使用HashSet。如果订单对您很重要，则至少需要一个LinkedHashSet。如果允许重复，请不要使用哈希值！使用LinkedList或数组。
顺便说一句，你分割字符串的方法没错，但会更复杂。你必须拆分：，然后调用str.trim（）删除任何额外的空格，str.substring（startIndx，endIndex），最后你可以解析数字列表。您必须使用str.indexOf（＆＃34; {＆＃34;）或手动搜索以获取索引。
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);
}
答案 1 :(得分：0)
最简单的方法是使用Set
http://docs.oracle.com/javase/7/docs/api/java/util/HashSet.html
以及Arrays.asList()
http://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html
将String[]转换为Set<String>：
Set<String> mySet = new HashSet<String>(Arrays.asList(s));
答案 2 :(得分：0)
这是一个有效的例子： 首先，您将创建一个正则表达式模式以匹配{}的内部，然后您将检查{}天气内部是否正确格式化。 然后，您将{}的内部转换为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();
}
}