Most efficient way to get the last element of a stream

Question

Stream doesn't have a last() method:

Stream<T> stream; T last = stream.last(); // No such method 

What's the most elegant and/or efficient way to get the last element (or null for an empty Stream)?

Answer 1

Do a reduction that simply returns the current value:

Stream<T> stream; T last = stream.reduce((a, b) -> b).orElse(null); 

Answer 2

This heavily depends on the nature of the Stream. Keep in mind that “simple” doesn’t necessarily mean “efficient”. If you suspect the stream to be very large, carrying heavy operations or having a source which knows the size in advance, the following might be substantially more efficient than the simple solution:

static <T> T getLast(Stream<T> stream) {     Spliterator<T> sp=stream.spliterator();     if(sp.hasCharacteristics(Spliterator.SIZED|Spliterator.SUBSIZED)) {         for(;;) {             Spliterator<T> part=sp.trySplit();             if(part==null) break;             if(sp.getExactSizeIfKnown()==0) {                 sp=part;                 break;             }         }     }     T value=null;     for(Iterator<T> it=recursive(sp); it.hasNext(); )         value=it.next();     return value; }  private static <T> Iterator<T> recursive(Spliterator<T> sp) {     Spliterator<T> prev=sp.trySplit();     if(prev==null) return Spliterators.iterator(sp);     Iterator<T> it=recursive(sp);     if(it!=null && it.hasNext()) return it;     return recursive(prev); } 

You may illustrate the difference with the following example:

String s=getLast(     IntStream.range(0, 10_000_000).mapToObj(i-> {         System.out.println("potential heavy operation on "+i);         return String.valueOf(i);     }).parallel() ); System.out.println(s); 

It will print:

potential heavy operation on 9999999 9999999 

In other words, it did not perform the operation on the first 9999999 elements but only on the last one.