TreeSet自定义比较器算法..字符串比较 [英] TreeSet Custom Comparator Algo .. String Comparision

问题描述

从提供的输入字符串中：

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}，

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

我要在TreeSet中添加相同的内容，并希望将其与第3个元素顺序进行排序，因此预期输出为：

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）

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

但是我的实际输出是：

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

(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11的字符串比较小于9，但以整数11> 9表示。我的预期输出变得不同。

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 + " ");

    }
}

推荐答案

您按字典顺序排序（ 123 在 20 之前）要做的就是将它们转换为整数，然后进行比较：

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:

否：

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

但是：

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

不过，一种更简洁的方法是创建一个包含这四个不同值的自定义对象，然后创建比较器，用于比较此类对象的3 ^rd 值：

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 3^rd value of such an object:

：

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);
    }
}

将打印：

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

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