TreeSet 自定义比较器算法 .. 字符串比较

发布于 2024-12-14 22:52:31 字数 1438 浏览 1 评论 0原文

从提供的输入字符串:

{ "200,400,7,1", "100,0,1,1", "200,200,3,1", "0,400,11,1", "407,308,5,1","100,600,9,1" } ,

我在 TreeSet 中添加相同的内容并希望它按第三个元素顺序排序,因此预期输出将是:

(100,0,1,1) (200,200,3,1) (407,308,5,1) (200,400,7,1) (100,600,9,1) (0,400,11,1)

但我的实际输出是:

(100,0,1,1)(0,400,11,1)(200,200,3,1)(407,308,5,1)(200,400,7,1)(100,600,9,1)

但是由于字符串比较11小于9,但就整数而言,11>9。我的预期输出越来越不同。建议我一些解决同样问题的想法。

import java.util.Comparator;
import java.util.TreeSet;

public class TreeSetComparator {

    public static void main(String[] args) {
        Comparator<String> comparator = new Comparator<String>() {
            @Override
            public int compare(String a, String b) {
                String aStr = a;
                String bStr = b;
                String[] splitA = aStr.split(",");
                String[] splitB = bStr.split(",");

                return splitA[2].compareTo(splitB[2]);
            }
        };

        String[] arr = { "200,400,7,1", "100,0,1,1", "200,200,3,1",
                "0,400,11,1", "407,308,5,1", "100,600,9,1" };

        TreeSet<String> ts = new TreeSet<String>(comparator);
        for (String str : arr) {
            ts.add(str);
        }

        for (String element : ts)
            System.out.print(element + " ");

    }
}

From the input string provided:

{ "200,400,7,1", "100,0,1,1", "200,200,3,1", "0,400,11,1",
"407,308,5,1","100,600,9,1" } ,

I am adding the same in a TreeSet and want it to be sorted with the 3rd element order, so the expected output will be:

(100,0,1,1) (200,200,3,1) (407,308,5,1) (200,400,7,1) (100,600,9,1) (0,400,11,1)

But my actual output is:

(100,0,1,1)(0,400,11,1)(200,200,3,1)(407,308,5,1)(200,400,7,1)(100,600,9,1)

But since the string comparison of 11 is less than 9 but in terms of integer , 11>9 . My expected output is getting differed. Suggest me some idea to resolve the same.

import java.util.Comparator;
import java.util.TreeSet;

public class TreeSetComparator {

    public static void main(String[] args) {
        Comparator<String> comparator = new Comparator<String>() {
            @Override
            public int compare(String a, String b) {
                String aStr = a;
                String bStr = b;
                String[] splitA = aStr.split(",");
                String[] splitB = bStr.split(",");

                return splitA[2].compareTo(splitB[2]);
            }
        };

        String[] arr = { "200,400,7,1", "100,0,1,1", "200,200,3,1",
                "0,400,11,1", "407,308,5,1", "100,600,9,1" };

        TreeSet<String> ts = new TreeSet<String>(comparator);
        for (String str : arr) {
            ts.add(str);
        }

        for (String element : ts)
            System.out.print(element + " ");

    }
}

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评论(1

素年丶 2024-12-21 22:52:31

您按照字典顺序排序("123" 位于 "20" 之前),您需要做的是将它们转换为整数,然后比较它们

: :

return splitA[2].compareTo(splitB[2]);

但是:

return Integer.valueOf(splitA[2]).compareTo(Integer.valueOf(splitB[2]));

但是,一种更简洁的方法是创建一个包含这 4 个不同值的自定义对象,然后创建一个比较器来比较该对象的第三个值:

以下内容:

public class Main {
    public static void main (String[] args) {

        Comparator<CustomObject> sortOn3rdValue = new Comparator<CustomObject>() {
            @Override
            public int compare(CustomObject o1, CustomObject o2) {
                return o1.v3 < o2.v3 ? -1 : o1.v3 > o2.v3 ? 1 : 0;
            }
        };

        Set<CustomObject> objects = new TreeSet<CustomObject>(sortOn3rdValue);

        String[] arr = { "200,400,7,1", "100,0,1,1", "200,200,3,1", "0,400,11,1", "407,308,5,1", "100,600,9,1" };

        for(String a : arr) {
            objects.add(new CustomObject(a.split(",")));
        }

        for(CustomObject co : objects) {
            System.out.println(co);
        }
    }
}

class CustomObject {

    final int v1, v2, v3, v4;

    CustomObject(String[] strValues) {
        // assume strValues.lenght == 4
        v1 = Integer.valueOf(strValues[0]);
        v2 = Integer.valueOf(strValues[1]);
        v3 = Integer.valueOf(strValues[2]);
        v4 = Integer.valueOf(strValues[3]);
    }

    @Override
    public String toString() {
        return String.format("(%d,%d,%d,%d)", v1, v2, v3, v4);
    }
}

会 打印:

(100,0,1,1)
(200,200,3,1)
(407,308,5,1)
(200,400,7,1)
(100,600,9,1)
(0,400,11,1)

You're sorting in a lexicographical order ("123" comes before "20"), what you need to do is convert them to integers, and then compare them:

Not:

return splitA[2].compareTo(splitB[2]);

but:

return Integer.valueOf(splitA[2]).compareTo(Integer.valueOf(splitB[2]));

However, a somewhat cleaner way would be to create a custom object holding these 4 different values and then create a Comparator that compares the 3rd value of such an object:

The following:

public class Main {
    public static void main (String[] args) {

        Comparator<CustomObject> sortOn3rdValue = new Comparator<CustomObject>() {
            @Override
            public int compare(CustomObject o1, CustomObject o2) {
                return o1.v3 < o2.v3 ? -1 : o1.v3 > o2.v3 ? 1 : 0;
            }
        };

        Set<CustomObject> objects = new TreeSet<CustomObject>(sortOn3rdValue);

        String[] arr = { "200,400,7,1", "100,0,1,1", "200,200,3,1", "0,400,11,1", "407,308,5,1", "100,600,9,1" };

        for(String a : arr) {
            objects.add(new CustomObject(a.split(",")));
        }

        for(CustomObject co : objects) {
            System.out.println(co);
        }
    }
}

class CustomObject {

    final int v1, v2, v3, v4;

    CustomObject(String[] strValues) {
        // assume strValues.lenght == 4
        v1 = Integer.valueOf(strValues[0]);
        v2 = Integer.valueOf(strValues[1]);
        v3 = Integer.valueOf(strValues[2]);
        v4 = Integer.valueOf(strValues[3]);
    }

    @Override
    public String toString() {
        return String.format("(%d,%d,%d,%d)", v1, v2, v3, v4);
    }
}

would print:

(100,0,1,1)
(200,200,3,1)
(407,308,5,1)
(200,400,7,1)
(100,600,9,1)
(0,400,11,1)
~没有更多了~
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