在Java中，根据值拆分字符串列表

Question

当字符串中的前6个字符为零("000000"(或字符串中的所有数字都为零时，我要做的是拆分字符串数组。限制为6个字符不会很动态。

我得到了这个代码，它实现了我想要实现的目标。

import java.util.*;
public class Main
{
public static void main(String[] args) {
ArrayList<String> unsplitted = new ArrayList<String>();

unsplitted.add("000000: this_should_go_into_first_array");
unsplitted.add("000234: something1");
unsplitted.add("0000ff: something2");
unsplitted.add("000111: something3");
unsplitted.add("000051: something4");
unsplitted.add("007543: something5");
unsplitted.add("000000: and_this_should_go_into_second_array");
unsplitted.add("005612: something7");
unsplitted.add("005712: something8");

System.out.println("Unsplitted list: "+ unsplitted);
List<String> arrlist1 = unsplitted.subList(0, 6);
List<String> arrlist2 = unsplitted.subList(6, unsplitted.size());

System.out.println("Sublist of arrlist1: "+ arrlist1);
System.out.println("Sublist of arrlist2: "+ arrlist2);
}
}

打印出想要的结果

Sublist of arrlist1: [000000: this_should_go_into_first_array, 000234: something1, 0000ff: something2, 000111: something3, 000051: something4, 007543: somethi
ng5]                                                                                                                                                          
Sublist of arrlist2: [000000: and_this_should_go_into_second_array, 005612: something7, 005712: something8]

然而，我事先不知道零的索引，那么如何通过动态查找零来获得相同的结果呢？

Answer 1

您可以简单地在数组中迭代并创建"桶"；每次检测到000000字符串时：

ArrayList<String> unsplitted = new ArrayList<String>();
unsplitted.add("000000: this_should_go_into_first_array");
unsplitted.add("000234: something1");
unsplitted.add("0000ff: something2");
unsplitted.add("000111: something3");
unsplitted.add("000051: something4");
unsplitted.add("007543: something5");
unsplitted.add("000000: and_this_should_go_into_second_array");
unsplitted.add("005612: something7");
unsplitted.add("005712: something8");
List<List<String>> results = new ArrayList<>();
unsplitted.forEach(w -> {
if(w.startsWith("000000") || results.isEmpty()) {
// no bucket or detect 000000
List<String> bucket = new ArrayList<>();
bucket.add(w);
results.add(bucket);
}
else {
// not contains 00000 put the value in the last bucket
results.get(results.size() - 1).add(w);
}
});
results.forEach(w -> {
System.out.println("Sublist " + w);
});

这是你所期望的结果吗？

结果：

Sublist [000000: this_should_go_into_first_array, 000234: something1, 0000ff: something2, 000111: something3, 000051: something4, 007543: something5]
Sublist [000000: and_this_should_go_into_second_array, 005612: something7, 005712: something8]

Answer 2

这个问题很有趣。有不同的方法可以实现这一点，但我将向您展示一个解决方案，它可以应用于第一部分的任何长度，我们可以将其视为关键
正如您在介绍中所说，如果检查仅限于6个字符，则不会是动态的。基于此，例如，您可以将字符"："的位置作为引用，并在数组的元素之间应用分区<这是我建议的解决方案：>

import java.util.*;
import java.util.function.*;
import java.util.stream.*;
public class Main
{
public static void main(String[] args) {
ArrayList<String> unsplitted = new ArrayList<String>();

unsplitted.add("000000: this_should_go_into_first_array");
unsplitted.add("000234: something1");
unsplitted.add("0000ff: something2");
unsplitted.add("000111: something3");
unsplitted.add("000051: something4");
unsplitted.add("007543: something5");
unsplitted.add("000000: and_this_should_go_into_second_array");
unsplitted.add("005612: something7");
unsplitted.add("005712: something8");
System.out.println("Non-split list: "+ unsplitted);
Predicate<String> filter = (String s) -> {
int indexOfCol = s.indexOf(":");
return s.substring(0, indexOfCol).equals("0".repeat(indexOfCol));
};
Map<Boolean, List<String>> splitMap = unsplitted.stream()
.collect(Collectors.partitioningBy(filter));
List<String> arrayZeroStart = splitMap.get(true);
List<String> arrayNonZeroStart = splitMap.get(false);
System.out.println("Sublist of arrayZeroStart: "+ arrayZeroStart);
System.out.println("Sublist of arrayWithout: "+ arrayNonZeroStart);
}
}

这是输出：

Non-split list: [000000: this_should_go_into_first_array, 000234: something1, 0000ff: 
something2, 000111: something3, 000051: something4, 007543: something5, 000000: and_this_should_go_into_second_array, 005612: something7, 005712: something8]
Sublist of arrayZeroStart: [000000: this_should_go_into_first_array, 000000: and_this_should_go_into_second_array]
Sublist of arrayWithout: [000234: something1, 0000ff: something2, 000111: something3, 000051: something4, 007543: something5, 005612: something7, 005712: something8]