你差点就到了。你已经有一对了，现在你所要做的就是在中间用“-”把它们打成一根绳子。
试一试：

Arrays.stream(exampleArray)
                .filter(s -> s.length() > 0)   // gets rid of blanks
                .filter(s -> !s.contains("junk"))
                .collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 2))
                .values()
                .stream()                        //stream the pairs
                .map(l -> String.join("-", l))   //and put a "-" between them & into a string
                .collect(Collectors.toList())    //collect all your joined String

实现此输出的简单方法是：

String a[] = {"a1", "junk", "a2", "b1", "junk", "b2", "c1", "junk", "junk", "junk", "c2", "d1", "junk", "d2", "junk-n"};

    Arrays.parallelSort(a); //sorting the array from a to z

    List<String> output = new ArrayList<>();

    for (int i = 0; i < a.length - 1; i++) {

        if(a[i].matches(".*\\d.*")) //verifying if have a number in string
           output.add(a[i] + " - " + a[++i]);

    }

    System.out.println(output); //["a1 - a2", "b1 - b2", "c1 - c2", "d1 - d2"]

collectors.groupingby有一个重载方法，您可以将结果传递给下游收集器。事实上，默认情况下，它使用了tolist收集器，所以您得到了list。可以使用joiningby连接字符串。

.collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 2
       , Collectors.joining(" - ")))

结果是

[a1-a2, b1-b2, c1-c2, d1-d2]

代码

public static void main(String[] args) {
    String exampleArray[] = new String[] {"a1", "junk", "a2", "b1", "junk", "b2", "c1", "junk", "junk", "junk", "c2", "d1", "junk", "d2", "junk-n"};
    AtomicInteger counter = new AtomicInteger(0);
    Collection<String> ans = (Arrays.asList(exampleArray)
            .stream()
            .filter(s -> s.length() > 0)   // gets rid of blanks
            .filter(s -> !s.contains("junk"))
            .collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 2, Collectors.joining(" - ")))
            .values());

    System.out.println(ans);
}