本文列舉了十個使用一行代碼即可獨立完成(不依賴其他代碼)的業務邏輯，主要依賴的是Java8中的Lambda和Stream等新特性以及try-with-resources、JAXB等。
1、對列表/數組中的每個元素都乘以2
<code>int
[] ia =range
(1
,10
).map
(i -> i *2
).toArray(); List result =range
(1
,10
).map
(i -> i *2
).boxed().collect(toList());/<code>
2、計算集合/數組中的數字之和
<code>range
(1
,1000
).sum
();range
(1
,1000
).reduce
(0
,Integer
::sum);Stream
.iterate
(0
, i -> i +1
).limit
(1000
).reduce
(0
,Integer
::sum);IntStream
.iterate
(0
, i -> i +1
).limit
(1000
).reduce
(0
,Integer
::sum);/<code>
3、驗證字符串是否包含集合中的某一字符串
<code>final
List<String
> keywords = Arrays.asList("brown"
,"fox"
,"dog"
,"pangram"
);final
String
tweet ="The quick brown fox jumps over a lazy dog. #pangram http://www.rinkworks.com/words/pangrams.shtml"
; keywords.stream().anyMatch(tweet::contains); keywords.stream().reduce(false
, (b, keyword) -> b || tweet.contains(keyword), (l, r) -> l || r);/<code>
4、讀取文件內容
原作者認為try with resources也是一種單行代碼編程。
<code>try
(BufferedReader reader =new
BufferedReader(new
FileReader("data.txt"
))) {String
fileText = reader.lines().reduce(""
,String
::concat); }try
(BufferedReader reader =new
BufferedReader(new
FileReader("data.txt"
))) { List<String
> fileLines = reader.lines().collect(toCollection(LinkedList<String
>::new
)); }try
(Stream<String
> lines = Files.lines(new
File("data.txt"
).toPath(), Charset.defaultCharset())) { List<String
> fileLines = lines.collect(toCollection(LinkedList<String
>::new
)); }/<code>
5、輸出歌曲《Happy Birthday to You!》 - 根據集合中不同的元素輸出不同的字符串
<code>range
(1
,5
).boxed().map
(i -> { out."Happy Birthday "
);if
(i ==3
)return
"dear NAME"
;else
return
"to You"
; }).forEach(out::println
);/<code>
6、過濾並分組集合中的數字
<code>Map
List>
result
=
Stream.of(49,
58
,
76
,
82
,
88
,
90
).collect(groupingBy(forPredicate(i
->
i
>
60
,
"passed"
,
"failed"
)));
/<code>
7、獲取並解析xml協議的Web Service
<code>FeedType feed = JAXB.unmarshal(new
URL("http://search.twitter.com/search.atom?&q=java8"
), FeedType.
class
); JAXB.marshal(feed, System.out);/<code>
8、獲得集合中最小/最大的數字
<code>int
min
=
Stream.of(14,
35
,
-7
,
46
,
98
).reduce(Integer::min).get();
min
=
Stream.of(14,
35
,
-7
,
46
,
98
).min(Integer::compare).get();
min
=
Stream.of(14,
35
,
-7
,
46
,
98
).mapToInt(Integer::new).min();
int
max
=
Stream.of(14,
35
,
-7
,
46
,
98
).reduce(Integer::max).get();
max
=
Stream.of(14,
35
,
-7
,
46
,
98
).max(Integer::compare).get();
max
=
Stream.of(14,
35
,
-7
,
46
,
98
).mapToInt(Integer::new).max();
/<code>
9、並行處理
<code>long
result = dataList.parallelStream().mapToInt(line -> processItem(line)).sum();/<code>
10、集合上的各種查詢(LINQ in Java)
<code>List
albums = Arrays.asList(unapologetic, tailgates, red); albums.stream() .filter(a -> a.tracks.stream().anyMatch(t -> (t.rating >=4
))) .sorted(comparing(album -> album.name)) .forEach
(album -> System.out.println(album.name));List
allTracks = albums.stream() .flatMap(album -> album.tracks.stream()) .collect(toList()); MapList> tracksByRating = allTracks.stream() .collect(groupingBy(Track::getRating));/<code>
