我有一个ArrayList
,每个索引有几个对象。我想特别用一个对象按字母顺序排列这个列表。对象是" my_id"此对象的值类似于:1A,10B,11B,2C,205Z等
我需要对它们进行排序:1A,2C,10B,11B,205Z。在首先对数字部分进行排序的情况下,将α部分排序为次要部分。 1,2,3,4,5,... A,B,C,D,E,......
我检查了一些非常有效的字母数字字符串排序: http://sanjaal.com/java/206/java-data-structure/alphanumeric-string-sorting-in-java-implementation/
不幸的是,我只能对该对象进行排序,因此我丢失了ArrayList
中的其他对象。我真的需要一种排序算法,可以通过我选择的对象重新排列ArrayList
索引,而不会丢失其他对象!
有没有办法在那里做到这一点?我一直找不到。我认为添加ArrayList
中的所有对象都是映射字符串非常有用:ArrayList<HashMap<String, String>>
[编辑]
我有我的阵列:
ArrayList<HashMap<String, String>> al
然后我存储对象:
String[] alphaNumericStringArray = new String[al.size()];
for(int i = 0; i < al.size(); i++)
{
alphaNumericStringArray[i] = al.get(i).get("my_id");
}
我现在对字符串数组进行排序:
// Sort the array now.
Arrays.sort(alphaNumericStringArray, new AlphanumericSorting());
然后我把对象放回去了:
for(int i = 0; i < al.size(); i++)
{
HashMap<String, String> map = new HashMap<String, String>();
map.put("my_id", alphaNumericStringArray[i]);
// TODO, need to append the rest of the objects.
al.set(i, map);
}
我知道你在想什么,我在重新映射时不会添加所有对象。这就是我目前所拥有的,但我想要的是一种对整个列表进行排序的方法,而不仅仅是一个对象&#34; my_id&#34;。我想重新安排索引,所以我不必在最后重新映射所有内容。
答案 0 :(得分:2)
运行main
方法:
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class Sorter {
public static void main(String[] args) {
List<String> unsorted = Arrays.asList("1A", "10B", "B", "753c", "Z", "M7", "32x", "11B", "2C", "205Z");
Collections.sort(unsorted, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
if (o1.isEmpty())
return -1;
if (o2.isEmpty())
return 1;
String o1number = extractNumberPrefix(o1);
String o2number = extractNumberPrefix(o2);
if (o1number.isEmpty())
if (o2number.isEmpty())
return o1.compareTo(o2);
else return 1;
if (o2number.isEmpty())
return -1;
if (o1number.equals(o2number))
return o1.compareTo(o2);
return Integer.parseInt(o1number) - Integer.parseInt(o2number);
}
private String extractNumberPrefix(String o1) {
String result = "";
for (int i = 0; i < o1.length(); i++) {
try {
Integer.parseInt(o1.substring(i, i + 1));
result += o1.substring(i, i + 1);
} catch (Exception e) {
break;
}
}
return result;
}
});
System.out.println("sorted = " + unsorted);
}
}
返回:
sorted = [1A, 2C, 10B, 11B, 32x, 205Z, 753c, B, M7, Z]
答案 1 :(得分:0)
经过仔细重建比较器和所有评论后,我终于想出了如何做到这一点。
<强>问题:强>
重申我的目标,以及解决方案。我有ArrayList<HashMap<String, String>>
。我想通过HashMap中的一个对象对ArrayList进行排序。我的HashMap中有多个对象,所以我想保留Array的整个索引。我还想按字母顺序排序,其中数值是第一个被排序的,而不是按字母顺序排序。即,1,2,3,4,...... A,B,C,D,......
<强>参考文献:强> http://sanjaal.com/java/206/java-data-structure/alphanumeric-string-sorting-in-java-implementation/
TL; DR解决方案:
在我的自定义Comparator函数public int compare(object firstObj, Object secondObj)
中,我需要将String值更改为HashMap对象引用/值。这里KEY_ID是我想要排序的对象。一旦我这样做,我使用Collections.sort按HashMap比较器排序,而不是Arrays.sort(集合处理ArrayList / HashMaps)。
代码解决方案:
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
/**
* DOCUMENTATION:
* http://sanjaal.com/java/206/java-data-structure/alphanumeric-string-sorting-in-java-implementation/
**/
@SuppressWarnings({"rawtypes", "unchecked"})
public class AlphanumericSorting implements Comparator
{
public int compare(Object firstObjToCompare, Object secondObjToCompare)
{
String firstString = ((HashMap<String,String>) firstObjToCompare).get("KEY_ID");
String secondString = ((HashMap<String,String>) secondObjToCompare).get("KEY_ID");
//String firstString = firstObjToCompare.toString();
//String secondString = secondObjToCompare.toString();
if (secondString == null || firstString == null)
{
return 0;
}
int lengthFirstStr = firstString.length();
int lengthSecondStr = secondString.length();
int index1 = 0;
int index2 = 0;
while(index1 < lengthFirstStr && index2 < lengthSecondStr)
{
char ch1 = firstString.charAt(index1);
char ch2 = secondString.charAt(index2);
char[] space1 = new char[lengthFirstStr];
char[] space2 = new char[lengthSecondStr];
int loc1 = 0;
int loc2 = 0;
do
{
space1[loc1++] = ch1;
index1++;
if (index1 < lengthFirstStr)
{
ch1 = firstString.charAt(index1);
}
else
{
break;
}
}
while (Character.isDigit(ch1) == Character.isDigit(space1[0]));
do
{
space2[loc2++] = ch2;
index2++;
if (index2 < lengthSecondStr)
{
ch2 = secondString.charAt(index2);
} else
{
break;
}
}
while (Character.isDigit(ch2) == Character.isDigit(space2[0]));
String str1 = new String(space1);
String str2 = new String(space2);
int result;
if (Character.isDigit(space1[0]) && Character.isDigit(space2[0]))
{
Integer firstNumberToCompare = new Integer(Integer.parseInt(str1.trim()));
Integer secondNumberToCompare = new Integer(Integer.parseInt(str2.trim()));
result = firstNumberToCompare.compareTo(secondNumberToCompare);
}
else
{
result = str1.compareTo(str2);
}
if (result != 0)
{
return result;
}
}
return lengthFirstStr - lengthSecondStr;
}
/**
* ALPHANUMERIC SORTING
*/
public static ArrayList<HashMap<String, String>> sortArrayList(ArrayList<HashMap<String, String>> al)
{
Collections.sort(al, new AlphanumericSorting());
return al;
}
}
返回已排序的ArrayList:
myArrayList = AlphanumericSorting.sortArrayList(myArrayList);
其中,
ArrayList<HashMap<String, String>> myArrayList;