You can use BinaryOperator<Integer> in this case like so :
BinaryOperator<Integer> add = (a, b) -> a + b;//lambda a, b : a + b
BinaryOperator<Integer> sub = (a, b) -> a - b;//lambda a, b : a - b
// Then create a new Map which take the sign and the corresponding BinaryOperator
// equivalent to signs = {"+" : lambda a, b : a + b, "-" : lambda a, b : a - b}
Map<String, BinaryOperator<Integer>> signs = Map.of("+", add, "-", sub);
int a = 5; // a = 5
int b = 3; // b = 3
// Loop over the sings map and apply the operation
signs.values().forEach(v -> System.out.println(v.apply(a, b)));
Outputs
8
2
Note for Map.of("+", add, "-", sub); I'm using Java 10, If you are not using Java 9+ you can add to your map like so:
Map<String, BinaryOperator<Integer>> signs = new HashMap<>();
signs.put("+", add);
signs.put("-", sub);
Ideone demo
Good practice
As already stated by @Boris the Spider and @Holger in the comments, Its better to use IntBinaryOperator to avoid boxing, in the end your code can look like this :
// signs = {"+" : lambda a, b : a + b, "-" : lambda a, b : a - b}
Map<String, IntBinaryOperator> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);
int a = 5; // a = 5
int b = 3; // b = 3
// for i in signs.keys(): print(signs[i](a,b))
signs.values().forEach(v -> System.out.println(v.applyAsInt(a, b)));