确定是否由Java 8中的anagram元素组成的列表 [英] Determine if a list composed of anagram elements in Java 8

问题描述

我想确定列表是否是使用Java 8的字谜。

I want to determine if a list is anagram or not using Java 8.

示例输入：

"cat", "cta", "act", "atc", "tac", "tca"

我写了以下函数来完成这项工作，但我想知道是否有更好更优雅的方法来做到这一点。

I have written the following function that does the job but I am wondering if there is a better and elegant way to do this.

boolean isAnagram(String[] list) {
    long count = Stream.of(list)
            .map(String::toCharArray)
            .map(arr -> {
                Arrays.sort(arr);
                return arr;
            })
            .map(String::valueOf)
            .distinct()
            .count();
    return count == 1;

}

似乎我无法用<$排序char数组c $ c> Stream.sorted（）方法，这就是我使用第二个map运算符的原因。如果有一些方法我可以直接在char流而不是Stream of char数组上操作，那也会有所帮助。

It seems I can't sort char array with Stream.sorted() method so that's why I used a second map operator. If there is some way that I can operate directly on char stream instead of Stream of char array, that would also help.

推荐答案

而不是创建和排序 char [] 或 int [] ，这不能内联，因此中断 你可以在字符串中获得字符中的字符，并在将它们转换为数组之前对它们进行排序。请注意，这是 IntSteam ，而 String.valueOf（int []）将包含数组的内存地址，这在这里不是很有用，所以在这种情况下最好使用 Arrays.toString 。

Instead of creating and sorting a char[] or int[], which can not be done inline and thus "breaks" the stream, you could get a Stream of the chars in the Strings and sort those before converting them to arrays. Note that this is an IntSteam, though, and String.valueOf(int[]) will include the array's memory address, which is not very useful here, so better use Arrays.toString in this case.

boolean anagrams = Stream.of(words)
        .map(String::chars).map(IntStream::sorted)
        .map(IntStream::toArray).map(Arrays::toString)
        .distinct().count() == 1;

当然，你也可以使用 map（s - > Arrays。 toString（s.chars（）。sorted（）。toArray（）））而不是四个地图的系列。不确定速度是否存在（显着）差异，这可能主要是品味问题。

Of course, you can also use map(s -> Arrays.toString(s.chars().sorted().toArray())) instead of the series of four maps. Not sure if there's a (significant) difference in speed, it's probably mainly a matter of taste.

此外，您可以使用 IntBuffer.wrap 使数组具有可比性，这应该比 Arrays.toString 快得多（感谢 Holger 。

Also, you could use IntBuffer.wrap to make the arrays comparable, which should be considerably faster than Arrays.toString (thanks to Holger in comments).

boolean anagrams = Stream.of(words)
        .map(s -> IntBuffer.wrap(s.chars().sorted().toArray()))
        .distinct().count() == 1;

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