1

I have an String Array i have fileterd from a database with output like this in Java

[ABCD XYZ M1210, 
ABCD XYZ M149, 
ABCD XYZ M5130, 
ABCD XYZ N1420, 
ABCD XYZ T11299, 
ABCD XYZ S11044]

Im looking to sort my array as follows:

[ABCD XYZ M149, 
ABCD XYZ M1210, 
ABCD XYZ M5130, 
ABCD XYZ N1420,  
ABCD XYZ S11044,
ABCD XYZ T11299]

and it is the last element i specifically want

 ==> String theStringIWant = myStringArray.get(myStringArray.size() - 1);

What i need to dois first sort the letter after "XYZ" alphabetically and then sort the numerically after that so for example ABCD XYZ M1210 < ABCD XYZ M5130 as 5130 is greater than 1210.

Any help here would be much appreciated

*Any referencing to suitable libraries in java etc

Cheers

4

3 回答 3

4 
Arrays.sort(array, new Comparator<String>() {
    @Override
    public int compare(String s1, String s2) {
        String stringPart1 = extractStringPart(s1); 
        String stringPart2 = extractStringPart(s2); 
        int result = stringPart1.compare(stringPart2);
        if (result == 0) {
            int intPart1 = extractIntPart(s1);
            int intPart2 = extractIntPart(s2);
            result = Integer.compare(intPart1, intPart2);
        }
        return result;
    }
});

这两种提取方法留作练习。阅读 String 和/或 Matcher 的 javadoc 以了解如何操作。

于 2013-11-14T17:11:20.423 回答
2

您可以使用Collections.sort对内容进行排序。

同时，编写一个比较器进行比较，例如，

1 写一个比较器

public class MyComparator implements Comparator {

@Override
public int compare(Object o1, Object o2) {

    String s1 = (String)o1;
    String s2 = (String)o2;
    //String part has 10 characters, It is fixed.
    String strPart1 = s1.substring(0,10); 
    int intPart1 = Integer.valueOf(s1.substring(10));

    String strPart2 = s2.substring(0,10);
    int intPart2 = Integer.valueOf(s2.substring(10));

    int strCompareResult = strPart1.compareTo(strPart2);
    if(0 == strCompareResult )
    {
        return intPart1 - intPart2;
    }
    else
    {
        return strCompareResult;
    }
}

}

2 使用Collections.sort和比较器完成排序

    List<String> results = Arrays.asList("ABCD XYZ M1210", "ABCD XYZ M149",
            "ABCD XYZ M5130", "ABCD XYZ N1420", "ABCD XYZ T11299",
            "ABCD XYZ S11044");

    System.out.println("Before sorting... ...");
    System.out.println(results);

    System.out.println("After sorting... ... ");
    Collections.sort(results, new MyComparator());
    System.out.println(results);

控制台中的输出：

排序前......

[ABCD XYZ M1210, ABCD XYZ M149, ABCD XYZ M5130, ABCD XYZ N1420, ABCD XYZ T11299, ABCD XYZ S11044]

排序后……

[ABCD XYZ M149, ABCD XYZ M1210, ABCD XYZ M5130, ABCD XYZ N1420, ABCD XYZ S11044, ABCD XYZ T11299]
于 2013-11-14T17:10:42.037 回答
1

这是我特别想要的最后一个元素

以下程序首先提取字符串元素的最后一个整数部分。然后通过比较它们返回。

List<String> results = Arrays.asList("ABCD XYZ M1210", "ABCD XYZ M149",
        "ABCD XYZ M5130", "ABCD XYZ N1420", "ABCD XYZ T11299",
        "ABCD XYZ S11044");

        Collections.sort(results, new Comparator<String>(){

            @Override
            public int compare(String o1, String o2) {
                Integer x1 = Integer.parseInt(o1.substring(o1.lastIndexOf(" ")+2, o1.length()));
                Integer x2 = Integer.parseInt(o2.substring(o2.lastIndexOf(" ")+2, o2.length()));

                return x1.compareTo(x2);

            }
        });

        System.out.println(results);

输出：

[ABCD XYZ M149, 
ABCD XYZ M1210, 
ABCD XYZ N1420, 
ABCD XYZ M5130, 
ABCD XYZ S11044, 
ABCD XYZ T11299]
于 2013-11-14T17:21:06.047 回答