#6323 Java 8 version for Problem-Solving-Examples-in-RxJava

docs/Problem-Solving-Examples-in-RxJava.md

@@ -1,4 +1,4 @@
-This page will present some elementary RxJava puzzles and walk through some solutions (using the Groovy language implementation of RxJava) as a way of introducing you to some of the RxJava operators.
+This page will present some elementary RxJava puzzles and walk through some solutions as a way of introducing you to some of the RxJava operators.
 
 # Project Euler problem #1
 
@@ -7,10 +7,22 @@ There used to be a site called "Project Euler" that presented a series of mathem
 > If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.
 
 There are several ways we could go about this with RxJava.  We might, for instance, begin by going through all of the natural numbers below 1000 with [`range`](Creating-Observables#range) and then [`filter`](Filtering-Observables#filter) out those that are not a multiple either of 3 or of 5:
+### Java
+```java
+Observable<Integer> threesAndFives = Observable.range(1, 999).filter(e -> e % 3 == 0 || e % 5 == 0);
+```
+### Groovy
 ````groovy
 def threesAndFives = Observable.range(1,999).filter({ !((it % 3) && (it % 5)) });
 ````
 Or, we could generate two Observable sequences, one containing the multiples of three and the other containing the multiples of five (by [`map`](https://github.com/Netflix/RxJava/wiki/Transforming-Observables#map)ping each value onto its appropriate multiple), making sure to only generating new multiples while they are less than 1000 (the [`takeWhile`](Conditional-and-Boolean-Operators#takewhile-and-takewhilewithindex) operator will help here), and then [`merge`](Combining-Observables#merge) these sets:
+### Java
+```java
+Observable<Integer> threes = Observable.range(1, 999).map(e -> e * 3).takeWhile(e -> e < 1000);
+Observable<Integer> fives = Observable.range(1, 999).map(e -> e * 5).takeWhile(e -> e < 1000);
+Observable<Integer> threesAndFives = Observable.merge(threes, fives).distinct();
+```
+### Groovy
 ````groovy
 def threes = Observable.range(1,999).map({it*3}).takeWhile({it<1000});
 def fives = Observable.range(1,999).map({it*5}).takeWhile({it<1000});
@@ -21,6 +33,11 @@ Don't forget the [`distinct`](Filtering-Observables#distinct) operator here, oth
 Next, we want to sum up the numbers in the resulting sequence. If you have installed the optional `rxjava-math` module, this is elementary: just use an operator like [`sumInteger` or `sumLong`](Mathematical-and-Aggregate-Operators#suminteger-sumlong-sumfloat-and-sumdouble) on the `threesAndFives` Observable. But what if you don't have this module? How could you use standard RxJava operators to sum up a sequence and emit that sum?
 
 There are a number of operators that reduce a sequence emitted by a source Observable to a single value emitted by the resulting Observable. Most of the ones that are not in the `rxjava-math` module emit boolean evaluations of the sequence; we want something that can emit a number. The [`reduce`](Mathematical-and-Aggregate-Operators#reduce) operator will do the job:
+### Java
+```java
+Single<Integer> summer = threesAndFives.reduce(0, (a, b) -> a + b);
+```
+### Groovy
 ````groovy
 def summer = threesAndFives.reduce(0, { a, b -> a+b });
 ````
@@ -38,6 +55,12 @@ Here is how `reduce` gets the job done. It starts with 0 as a seed. Then, with e
  </tbody>
 </table>
 Finally, we want to see the result. This means we must [subscribe](Observable#onnext-oncompleted-and-onerror) to the Observable we have constructed:
+
+### Java
+```java
+summer.subscribe(System.out::print);
+```
+### Groovy
 ````groovy
 summer.subscribe({println(it);});
 ````
@@ -47,11 +70,24 @@ summer.subscribe({println(it);});
 How could you create an Observable that emits [the Fibonacci sequence](http://en.wikipedia.org/wiki/Fibonacci_number)?
 
 The most direct way would be to use the [`create`](Creating-Observables#wiki-create) operator to make an Observable "from scratch," and then use a traditional loop within the closure you pass to that operator to generate the sequence. Something like this:
+### Java
+```java
+Observable<Integer> fibonacci = Observable.create(emitter -> {
+    int f1 = 0, f2 = 1, f = 1;
+    while (!emitter.isDisposed()) {
+        emitter.onNext(f);
+        f = f1 + f2;
+        f1 = f2;
+        f2 = f;
+    }
+});
+```
+### Groovy
 ````groovy
-def fibonacci = Observable.create({ observer ->
-  def f1=0; f2=1, f=1;
-  while(!observer.isUnsubscribed() {
-    observer.onNext(f);
+def fibonacci = Observable.create({ emitter ->
+  def f1=0, f2=1, f=1;
+  while(!emitter.isDisposed()) {
+    emitter.onNext(f);
     f  = f1+f2;
     f1 = f2;
     f2 = f;
@@ -61,7 +97,16 @@ def fibonacci = Observable.create({ observer ->
 But this is a little too much like ordinary linear programming. Is there some way we can instead create this sequence by composing together existing Observable operators?
 
 Here's an option that does this:
-````
+### Java
+```java
+Observable<Integer> fibonacci =
+        Observable.fromArray(0)
+                  .repeat()
+                  .scan(new int[]{0, 1}, (a, b) -> new int[]{a[1], a[0] + a[1]})
+                  .map(a -> a[1]);
+```
+### Groovy
+````groovy
 def fibonacci = Observable.from(0).repeat().scan([0,1], { a,b -> [a[1], a[0]+a[1]] }).map({it[1]});
 ````
 It's a little [janky](http://www.urbandictionary.com/define.php?term=janky). Let's walk through it:
@@ -73,6 +118,11 @@ This has the effect of emitting the following sequence of items: `[0,1], [1,1],
 The second item in this array describes the Fibonacci sequence. We can use `map` to reduce the sequence to just that item.
 
 To print out a portion of this sequence (using either method), you would use code like the following:
+### Java
+```java
+fibonacci.take(15).subscribe(System.out::println);
+```
+### Groovy
 ````groovy
 fibonnaci.take(15).subscribe({println(it)})];
 ````