使用Java 8,创建排序和分组字符串列表的最简洁方法是什么

Question

使用Java 8,创建排序和分组的字符串列表的最简洁方法是什么？使用Lambdas和Collections and Streams框架显示旧方法和新方法.

您可以使用旧的(或新的)方式使用第三方库(流行的).

但是,我建议使用vanilla Java,因为它显示了Java 8中的语言更改带来的任务更改表.

Input: List<String>
Output: Map<Character<List<String>>
The key of map is 'A' to 'Z'
Each list in the map are sorted.

它将被分类和分组,以便......

鉴于此列表:"啤酒","苹果","香蕉","凤梨","芒果","蓝莓"

甲Map意愿产生包含第一信作为密钥.地图中的值将List是以该键(字母)开头的所有单词的排序:

• 关键:A值:["Ananas","Apple"]
• 关键:B值:["香蕉","啤酒","蓝莓"]
• 关键:M值:["芒果"]

Answer 1

使用Java,没有第三方库的帮助,有旧方法和新方法.只需使用Collections.sort(..)进行排序.

旧方法面临的挑战是需要大量代码来对值进行分组.

 - Input: List<String>
 - Output: Map<Character,<List<String>>
 - The key of map is 'A' to 'Z'
 - Each list in the map are sorted.

老Java

List<String> keywords = Arrays.asList("Apple", "Ananas", "Mango", "Banana", "Beer"); 
Map<Character, List<String>> result = new HashMap<Character, List<String>>(); 
for(String k : keywords) {   
    char firstChar = k.charAt(0);     
    if(!result.containsKey(firstChar)) {     
        result.put(firstChar, new  ArrayList<String>());   
    }     
    result.get(firstChar).add(k); 
} 
for(List<String> list : result.values()) {   
    Collections.sort(list); 
}
System.out.println(result); 

新Java 8

List<String> keywords = Arrays.asList("Apple", "Ananas", "Mango", "Banana", "Beer");

Map<Character, List<String>> result = keywords.stream()
     .sorted()
     .collect(Collectors.groupingBy(it -> it.charAt(0)));

System.out.println(result);

新的Java 8源数据已经作为'流'

正如@KevinO所说

 Map<Character, List<String>> result = Stream
      .of( "Apple", "Ananas", "Mango", "Banana","Beer")
      .sorted()
      .collect(Collectors.groupingBy(it -> it.charAt(0)))

System.out.println(result);

Answer 2

使用流行的第三方Guava库,与Java 6兼容:

TreeMultimap<Character, String> multimap = TreeMultimap.create();
for (String string : list) {
  multimap.put(string.charAt(0), string);
}
return Multimaps.asMap(ImmutableListMultimap.copyOf(multimap));

这会对重复字符串进行重复数据删除,因此允许重复字符串的备用版本:

ImmutableListMultimap.Builder<Character, String> builder = 
  ImmutableListMultimap.builder();
for (String string : Ordering.natural().sortedCopy(list)) {
  builder.put(string.charAt(0), string);
}
return Multimaps.asMap(builder.build());