如何将列表分发到子列表中，保持元素的原始顺序？ [英] How to distribute a list into sub-lists, keeping the original order of the elements?

问题描述

如何将列表拆分为给定数量的列表，按顺序获取元素并将它们分发到子列表（所以不对列表进行分区）？

How to split a list into a given number of lists, taking the elements in order and distributing them to the sub-lists (so not partitioning the list)?

我想尽可能好（使用Java 8功能或Guava或类似的东西。）

I would like to do this as "nice" as possible (using Java 8 features or Guava or something similar.

• 示例列表： [1 2 3 4 5 6 7]


• 应分为3： [1 4 7] [2 5] [3 6]


• 应分为2： [1 3 5 7] [2 4 6]

• Example list: [1 2 3 4 5 6 7]
• Should be split in 3 : [1 4 7] [2 5] [3 6]
• Should be split in 2 : [1 3 5 7] [2 4 6]

推荐答案

如果源列表支持有效的随机访问，例如 ArrayList 是的，你可以使用

If the source list supports efficient random access, like ArrayList does, you can use

IntStream.range(0, source.size()).boxed()
  .collect(groupingBy(i->i%listCount, LinkedHashMap::new, mapping(source::get, toList())));

例如

List<Integer> source=IntStream.range(0, 20).boxed().collect(toList());
System.out.println(source);
int listCount=5;

Map<Integer, List<Integer>> collect = IntStream.range(0, source.size()).boxed()
  .collect(groupingBy(i->i%listCount, LinkedHashMap::new, mapping(source::get, toList())));
// in case it really has to be a List:
List<List<Integer>> result=new ArrayList<>(collect.values());

result.forEach(System.out::println);

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[0, 5, 10, 15]
[1, 6, 11, 16]
[2, 7, 12, 17]
[3, 8, 13, 18]
[4, 9, 14, 19]

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