[英]Sort array first by length then alphabetically in Java
How can I sort an array first by length, then alphabetically?如何首先按长度对数组进行排序,然后按字母顺序排序?
I have a list of things with numbers on them, and I am currently getting:我有一个带有数字的东西清单,我目前得到:
Something1 Something10 Something2 Something3某物1 某物10 某物2 某物3
Whereas I want to get:而我想得到:
Something1 Something2 Something3 Something10某物1 某物2 某物3 某物10
public class MyComparator implements Comparator<String>{
@Override
public int compare(String o1, String o2) {
if (o1.length() > o2.length()) {
return 1;
} else if (o1.length() < o2.length()) {
return -1;
}
return o1.compareTo(o2);
}
}
Then use:然后使用:
Collections.sort(yourList, new MyComparator());
Here's a concise Java 8 solution:这是一个简洁的 Java 8 解决方案:
List<String> list = Arrays.asList("Something1", "Something10", "Something2", "Something3");
list.sort(Comparator.comparing(String::length).thenComparing(String::compareTo));
Or, case-insensitive version:或者,不区分大小写的版本:
list.sort(Comparator.comparing(String::length).thenComparing(String::compareToIgnoreCase));
创建一个 Comparator,首先按长度进行比较,如果长度相同,则使用 String.compareTo()。
Sorting first by length and then lexically will work ONLY if the string prefixes (ie the part before the number) is the same length in all cases.仅当字符串前缀(即数字之前的部分)在所有情况下长度相同时,首先按长度排序,然后按词法排序才有效。 I believe you may really want to write a comparator that separates the string and numeric parts and sorts alphabetically on the string and numerically on the number part.我相信您可能真的想编写一个比较器来分隔字符串和数字部分,并在字符串上按字母顺序排序,在数字部分上按数字排序。
By using lambda, The java program looking like.通过使用 lambda,java 程序看起来像。
import java.util.Arrays;
import java.util.List;
public class Test1 {
public static void main(String[] args) {
String str = "This is am example";
List<String> strl = Arrays.asList(str.split(" "));
strl.sort( (o1,o2)->{
if(o1.length() > o2.length()) return 1;
else if(o1.length() < o2.length()) return -1;
else return o1.compareTo(o2);
});
System.out.println(strl);
}
}
Define a class to hold your item in. Seems like you want it to be a String.定义一个类来保存您的项目。似乎您希望它是一个字符串。
For that class, you need to define the Comparable interface and put the logic to compare in its abstract method.对于该类,您需要定义 Comparable 接口并将要比较的逻辑放在其抽象方法中。
int compareTo(T o)
For example:例如:
class MyString extends String { @Override int compareTo(Object obj) { // put your logic in here. // Return -1 if this is "less than" obj. // Return 0 if this is equal to obj // Return 1 if this is "greater than" obj. // Test length first if (length() < obj.length()) return -1; if (length() > obj.length()) return 1; // Lengths are the same, use the alphabetical compare defined by String already return super.compareTo(obj); } }
Disclaimer, I didn't actually test this code, but it should be close to what you want.免责声明,我实际上并没有测试这段代码,但它应该接近你想要的。
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