java8 stream根据下标分组

Question

我有个list想要按照奇偶下标生成两组list，只利用java8的stream可以做到吗？
比如[1,3,5,7,9]，按照奇偶下标分成[1,5,9]和[3,7],但是stream似乎无法获取下标，所以试了半天也不行。
另外，java8的stream以后可以完全取代for循环吗？我感觉有些复杂的逻辑拿stream折腾不是一个好的选择

Answer 1

[1, 3, 5, 7, 9]的偶数下标数组：    int[] i = new int[] {0};     System.out.println(Arrays.toString(IntStream.of(1, 3, 5, 7, 9).filter(n -> i[0]++ % 2 == 0).toArray()));得到[1, 5, 9]

Answer 2

这个要从为什么java8会弄出stream这种循环方式说起了以前的for循环是属于外部循环，由调用者自己显式的取出每个元素加以处理，尤其是并行处理时，非常痛苦而stream是属于内部循环，你可以无需关心它是如何循环的，要是并行的时候如何优化，这些你都可以不管，你只需要简简单单告诉他需要做什么处理即可举个简单的例子：地上有很多乒乓球，爸爸要让你儿子捡起来放到盒子里如果是外部循环，可能这种模式爸爸：地上还有球么儿子：有爸爸：那把它捡起来放到盒子里，还有么？儿子：有爸爸：继续捡起来放到盒子里，还有么？儿子：有爸爸：接着捡起来放到盒子里，还有么？儿子：没有了爸爸：ok，收拾好了如果是内部循环爸爸：把地上的球捡起来全部放到盒子里儿子：好，完成例子可能很牵强，但是我们作为开发者，就是爸爸，只需要告诉儿子（JDK）把球放进盒子这个动作就好，具体儿子怎么放，一次一个手拿一个，还是一次每个手拿一个，由它自己来思考，不需要我们去指挥，我们只关注具体需要做的事就好所以说stream是一种内部循环，你不需要去关注下标...关注下标的话，从某种意义上来说，还是外部循环的思想在处理了，当然题主的问题中，明确就是要求是关注下标了，貌似好像stream是没法写似的其实不然，写肯定能写的，只是说换一种方式而已，不把下标当成循环用的元素，而是把它与具体需要循环的数字绑定起来，写出来差不多是这个样子public static void main(String[] args) {&nbsp; &nbsp; &nbsp; &nbsp; List<Integer> list = Arrays.asList(1,3,5,7,9);&nbsp; &nbsp; &nbsp; &nbsp; Map<Boolean, List<Integer>> collect = IntStream.rangeClosed(1, list.size())&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .boxed()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // 把数字和下标绑定起来&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .map(i -> mapToEntry(i, list.get(i-1)))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // 按照下标是否能被2整除进行分组&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(Collectors.partitioningBy(simpleEntry -> Integer.parseInt(simpleEntry.getKey().toString()) % 2 == 0,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Collectors.mapping(Map.Entry::getValue, Collectors.toList())));&nbsp; &nbsp; &nbsp; &nbsp; // 偶数列表 3,9&nbsp; &nbsp; &nbsp; &nbsp; List<Integer> evenList = collect.get(Boolean.TRUE);&nbsp; &nbsp; &nbsp; &nbsp; // 奇数列表 1,5,7&nbsp; &nbsp; &nbsp; &nbsp; List<Integer> oddList = collect.get(Boolean.FALSE);&nbsp; &nbsp; }&nbsp; &nbsp; private static AbstractMap.SimpleEntry<Integer, Integer> mapToEntry(Integer i, Integer integer) {&nbsp; &nbsp; &nbsp; &nbsp; return new AbstractMap.SimpleEntry(i, integer);&nbsp; &nbsp; }