我正在创建一个目录，而我所拥有的是产品编号到页面的地图。因此，条目可能如下所示：
ABC123 => [59, 58, 57, 19, 36, 15, 33, 34, 13, 39, 11, 37, 38, 21, 20, 40, 63, 60, 45, 46, 22, 23, 24, 26, 3, 2, 10, 1, 7, 6, 5, 4, 8]
我想从中得到的是：
1-8,10,11,13,15,19-24,26,33,34,36-38,40,45,46,57-60
我当然可以编写代码，但我认为其他人已经解决了这个问题。我的谷歌搜索没有任何结果。
我一如既往地感谢您提供的任何帮助！
答案 0 :(得分：4)
您可以将数字收集到有序集中，然后迭代数字。
快速而肮脏的例子：
SortedSet<Integer> numbers = new TreeSet<Integer>();
numbers.add( 1 );
numbers.add( 2 );
numbers.add( 3 );
numbers.add( 6 );
numbers.add( 7 );
numbers.add( 10 );
Integer start = null;
Integer end = null;
for( Integer num : numbers ) {
//initialize
if( start == null || end == null ) {
start = num;
end = num;
}
//next number in range
else if( end.equals( num - 1 ) ) {
end = num;
}
//there's a gap
else {
//range length 1
if( start.equals( end )) {
System.out.print(start + ",");
}
//range length 2
else if ( start.equals( end - 1 )) {
System.out.print(start + "," + end + ",");
}
//range lenth 2+
else {
System.out.print(start + "-" + end + ",");
}
start = num;
end = num;
}
}
if( start.equals( end )) {
System.out.print(start);
}
else if ( start.equals( end - 1 )) {
System.out.print(start + "," + end );
}
else {
System.out.print(start + "-" + end);
}
收益率：1-3,6,7,10
答案 1 :(得分：3)
Apache Commons具有您可以使用的IntRange类型。不幸的是，我没有找到一组相应的实用程序来创建它们。以下是您可以使用的基本方法：
//create a list of 1-integer ranges
List<IntRange> ranges = new LinkedList<IntRange>();
for ( int pageNum : pageNums ) {
ranges.add(new IntRange(pageNum));
}
//sort the ranges
Collections.sort(ranges, new Comparator<IntRange>() {
public int compare(IntRange a, IntRange b) {
return Integer.valueOf(a.getMinimumInteger()).compareTo(b.getMinimumInteger());
}
});
List<IntRange> output = new ArrayList<IntRange>();
if ( ranges.isEmpty() ) {
return output;
}
//collapse consecutive ranges
IntRange range = ranges.remove(0);
while ( !ranges.isEmpty() ) {
IntRange nextRange = ranges.remove(0);
if ( range.getMaximumInteger() == nextRange.getMinimumInteger() - 1 ) {
range = new IntRange(range.getMinimumInteger(), nextRange.getMaximumInteger());
} else {
output.add(range);
range = nextRange;
}
}
output.add(range);
或者，您可以跳过第一步，直接从已排序的页码列表中创建范围。
答案 2 :(得分：2)
修改：更好的描述：
我必须处理类似于支持有限范围的有序集合的东西，我使用Google的Guava Range类和二分搜索的混合来在相应的范围插入元素或创建一个新的单例范围（范围为1 element），最终有更多插入，范围有扩展的机会（或者在删除时缩小/拆分），删除非常快，因为找到元素使用二进制搜索的相应范围：
import com.google.common.collect.DiscreteDomains;
import com.google.common.collect.Lists;
import com.google.common.collect.Range;
import com.google.common.collect.Ranges;
import java.util.Collection;
import java.util.List;
public class IntRangeCollection
{
private int factor=10;
private List<Range<Integer>> rangeList=null;
public IntRangeCollection()
{
rangeList=Lists.newArrayListWithExpectedSize(1000);
}
public IntRangeCollection(final int size)
{
rangeList=Lists.newArrayListWithExpectedSize(size);
}
public IntRangeCollection(final int size, final int factor)
{
rangeList=Lists.newArrayListWithExpectedSize(size);
this.factor=factor;
}
protected IntRangeCollection(final List<Range<Integer>> rangeList)
{
this.rangeList=rangeList;
}
public static IntRangeCollection buildIntRangesCollectionFromArrays(final List<Integer[]> arrays)
{
final List<Range<Integer>> rangeList=Lists.newArrayListWithCapacity(arrays.size());
for(Integer[] range : arrays){
rangeList.add(range.length == 1 ? Ranges.singleton(range[0]) : Ranges.closed(range[0], range[1]));
}
return new IntRangeCollection(rangeList);
}
public boolean addElements(final Collection<Integer> elements)
{
boolean modified=false;
for(Integer element : elements){
modified=addElement(element) || modified;
}
return modified;
}
public boolean removeElements(final Collection<Integer> elements)
{
boolean modified=false;
for(Integer element : elements){
modified=removeElement(element) || modified;
}
return modified;
}
public boolean addElement(final Integer element)
{
final Range<Integer> elementRange=Ranges.singleton(element);
if(rangeList.isEmpty()){
rangeList.add(elementRange);
} else{
int
start=0, mid=0,
end=rangeList.size() - 1;
Range<Integer> midRange=null;
while(start<=end){
mid=(start + end) / 2;
midRange=rangeList.get(mid);
if(midRange.contains(element)){
return false;
} else if(testLinkable(midRange, element)){
rangeList.set(mid, midRange.span(elementRange));
if(mid>0){
final Range<Integer> a=rangeList.get(mid - 1);
if(testLinkable(a, midRange)){
rangeList.set(mid - 1, a.span(midRange));
rangeList.remove(mid);
mid--;
}
}
if(mid<rangeList.size() - 1){
final Range<Integer> b=rangeList.get(mid + 1);
if(testLinkable(midRange, b)){
rangeList.set(mid, midRange.span(b));
rangeList.remove(mid + 1);
}
}
return true;
} else if(midRange.lowerEndpoint().compareTo(element)<0){
start=mid + 1;
} else{
end=mid - 1;
}
}
//noinspection ConstantConditions
rangeList.add(midRange.lowerEndpoint().compareTo(element)<0 ? mid + 1 : mid, elementRange);
}
return true;
}
public boolean removeElement(final Integer element)
{
final Range<Integer> elementRange=Ranges.singleton(element);
if(rangeList.isEmpty()){
rangeList.add(elementRange);
} else{
int
start=0, mid,
end=rangeList.size() - 1;
while(start<=end){
mid=(start + end) / 2;
final Range<Integer> midRange=rangeList.get(mid);
if(midRange.contains(element)){
final Integer
lower=midRange.lowerEndpoint(),
upper=midRange.upperEndpoint();
if(lower.equals(upper)){
rangeList.remove(mid);
} else if(lower.equals(element)){
rangeList.set(mid, Ranges.closed(element + 1, upper));
} else if(upper.equals(element)){
rangeList.set(mid, Ranges.closed(lower, element - 1));
} else{
rangeList.set(mid, Ranges.closed(element + 1, upper));
rangeList.add(mid, Ranges.closed(lower, element - 1));
}
return true;
} else if(midRange.lowerEndpoint().compareTo(element)<0){
start=mid + 1;
} else{
end=mid - 1;
}
}
}
return false;
}
public List<Integer> getElementsAsList()
{
final List<Integer> result=Lists.newArrayListWithExpectedSize(rangeList.size() * factor);
for(Range<Integer> range : rangeList){
result.addAll(range.asSet(DiscreteDomains.integers()));
}
return result;
}
public List<Integer[]> getRangesAsArray()
{
final List<Integer[]> result=Lists.newArrayListWithCapacity(rangeList.size());
for(Range<Integer> range : rangeList){
final Integer
lower=range.lowerEndpoint(),
upper=range.upperEndpoint();
result.add(lower.equals(upper) ? new Integer[]{lower} : new Integer[]{lower,upper});
}
return result;
}
public int getRangesCount()
{
return rangeList.size();
}
private boolean testLinkable(final Range<Integer> range, final Integer element)
{
return Ranges.closed(range.lowerEndpoint() - 1, range.upperEndpoint() + 1).contains(element);
}
private boolean testLinkable(final Range<Integer> a, final Range<Integer> b)
{
return Ranges.closed(a.lowerEndpoint() - 1, a.upperEndpoint() + 1).isConnected(b);
}
@Override
public String toString()
{
return "IntRangeCollection{" +
"rangeList=" + rangeList +
'}';
}
public static void main(String[] args)
{
final int MAX_NUMBER=1000;
final long startMillis=System.currentTimeMillis();
final IntRangeCollection ranges=new IntRangeCollection();
for(int i=0; i<MAX_NUMBER; i++){
//noinspection UnsecureRandomNumberGeneration
ranges.addElement((int) (Math.random() * MAX_NUMBER));
}
System.out.println(MAX_NUMBER + " contained in " + ranges.rangeList.size() + " ranges done in " + (System.currentTimeMillis() - startMillis) + "ms");
System.out.println(ranges);
for(int i=0; i<MAX_NUMBER / 4; i++){
//noinspection UnsecureRandomNumberGeneration
ranges.removeElement((int) (Math.random() * MAX_NUMBER));
}
System.out.println(MAX_NUMBER + " contained in " + ranges.rangeList.size() + " ranges done in " + (System.currentTimeMillis() - startMillis) + "ms");
System.out.println(ranges);
}
}
答案 3 :(得分：1)
您可以使用Arrays.sort（）并查找相邻的重复项/范围。但是我怀疑TreeSet可能更容易使用。
答案 4 :(得分：1)
This is a good example，它显示了一种简单的方法来实现这一目标。