英文:
Java sort an array of strings (mixed with numbers and charcter)
问题
我正在Java中对一个数组进行排序，我的输入是{"a3"，"a2"，"a11"，"b1"，"b2"，"b3"，"c3"，"c13"，"c2"}，我希望输出是{"a2"，"a3"，"a11"，"b1"，"b2"，"b3"，"c2"，"c3"，"c13"}
以下是我在下面的代码中所做的，但没有返回正确的结果，任何建议/代码示例将不胜感激：
import java.util.Arrays;
public class Main
{
public static void main(String[] args) {
String[] var = {"a3", "a2", "a11", "b1", "b2", "b3", "c3", "c13", "c2"};
Arrays.sort(var, (s1, s2) -> {
String[] parts1 = s1.split("(?<=\\D)(?=\\d)");
String[] parts2 = s2.split("(?<=\\D)(?=\\d)");
int compare = parts1[0].compareTo(parts2[0]);
if (compare == 0) {
if (parts1.length > 1 && parts2.length > 1) {
return Integer.compare(Integer.parseInt(parts1[1]), Integer.parseInt(parts2[1]));
}
}
return compare;
});
System.out.println(Arrays.toString(var));
}
}
从上面的代码中，我得到的输出是[a2, a3, a11, b1, b2, b3, c2, c3, c13]
英文:
I'm sorting an array in Java, my inputs are {"a3", "a2", "a11", "b1", "b2", "b3", "c3", "c13", "c2"} and I want output {"a2", "a3", "a11", "b1", "b2", "b3", "c2", "c3", "c13"}
What I'm doing in below not returning proper results, any suggestions/code example appreciated
import java.util.Arrays;
public class Main
{
public static void main(String[] args) {
String[] var = {"a3", "a2", "a11", "b1", "b2", "b3", "c3", "c13", "c2"};
Arrays.sort(var);
System.out.println(Arrays.toString(var));
}
}
From above code I'm getting output [a11, a2, a3, b1, b2, b3, c13, c2, c3]
答案1
得分: 1
这里有一个相对简单的方法。然而，对于像这样的情况，我会创建一个类来分别保存字母部分和数字部分，并编写一个比较器。在打印这些对象时，我只需使 toString() 返回原始值。
这是基础方法。它不会检查格式错误的字符串。
String[] var = {"a3", "a2", "a11", "b1", "b2", "b3", "c3", "c13", "c2"};
Comparator<String> comp = (a,b) -> {
// 在最后一个字符和第一个数字之间分割字符串
String[] v1 = a.split("(?<![\\d])(?=\\d)");
String[] v2 = b.split("(?<![\\d])(?=\\d)");
// 将每个整数部分转换为整数。
int n1 = Integer.valueOf(v1[1]);
int n2 = Integer.valueOf(v2[1]);
// 比较两者并获取结果
int r1 = v1[0].compareTo(v2[0]);
int r2 = Integer.compare(n1, n2);
// 首先根据 r1（字符串）排序。如果它们相等（r1 == 0）
// 则根据比较整数的结果进行排序。
return r1 == 0 ? r2 : r1;
};
Array.sort 不接受比较器，所以你需要将数组转换为列表并在其上进行排序。这也会改变数组，因为对象数组会备份从 Arrays.asList 返回的列表。
Collections.sort(Arrays.asList(var), comp);
System.out.println(Arrays.toString(var));
打印结果：
[a2, a3, a11, b1, b2, b3, c2, c3, c13]
英文:
Here's a relatively simple approach. However, for something like this I would create a class to hold the alpha and numeric parts separately and write a comparator. When I print the objects I would simply have toString() return the original value.
This is basic. It does not check for malformed strings.
String[] var = {"a3", "a2", "a11", "b1", "b2", "b3", "c3", "c13", "c2"};
Comparator<String> comp = (a,b)->{
// split the strings between the last char and the first digit
String[] v1 = a.split("(?<!\\d)(?=\\d)");
String[] v2 = b.split("(?<!\\d)(?=\\d)");
// convert each integer part to an int.
int n1 = Integer.valueOf(v1[1]);
int n2 = Integer.valueOf(v2[1]);
// compare each and get the result
int r1 = v1[0].compareTo(v2[0]);
int r2 = Integer.compare(n1,n2);
// first sort on r1 (the strings). If they are equal ( r1 == 0)
// then sort on the result of comparing the integers.
return r1 == 0 ? r2 : r1;
};
Array.sort does not take a comparator so you need to convert the array to a list and sort on that. This will also change the array since object arrays back up the returned list from Arrays.asList.
Collections.sort(Arrays.asList(var),comp);
System.out.println(Arrays.toString(var));
Prints
[a2, a3, a11, b1, b2, b3, c2, c3, c13]
</details>
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论