Answer by John Fowler for How to sort List of objects by some property

JAVA 8 and Above Answer (Using Lambda Expressions)

In Java 8, Lambda expressions were introduced to make this even easier! Instead of creating a Comparator() object with all of it's scaffolding, you can simplify it as follows: (Using your object as an example)

Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.timeStarted-a2.timeStarted);

or even shorter:

Collections.sort(list, Comparator.comparingInt(ActiveAlarm ::getterMethod));

That one statement is equivalent to the following:

Collections.sort(list, new Comparator<ActiveAlarm>() {    @Override    public int compare(ActiveAlarm a1, ActiveAlarm a2) {        return a1.timeStarted - a2.timeStarted;    }});

Think of Lambda expressions as only requiring you to put in the relevant parts of the code: the method signature and what gets returned.

Another part of your question was how to compare against multiple fields. To do that with Lambda expressions, you can use the .thenComparing() function to effectively combine two comparisons into one:

Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.timeStarted-a2.timeStarted                    .thenComparing ((ActiveAlarm a1, ActiveAlarm a2) -> a1.timeEnded-a2.timeEnded));

The above code will sort the list first by timeStarted, and then by timeEnded (for those records that have the same timeStarted).

One last note: It is easy to compare 'long' or 'int' primitives, you can just subtract one from the other. If you are comparing objects ('Long' or 'String'), I suggest you use their built-in comparison. Example:

Collections.sort(list, (ActiveAlarm a1, ActiveAlarm a2) -> a1.name.compareTo(a2.name) );

EDIT: Thanks to Lukas Eder for pointing me to .thenComparing() function.