Technical interview question: F (f (n)) = =-N

Recently encountered a technical interview question, here to share. The title is to design a function f that makes

F (f (n)) =-N

Here n is a 32-bit integer. You cannot use imaginary or plural operations.

If you can't design a function to make it work for all integers under 32 bits, design this function so that it can be applied to as many integers as possible.

Design a function f, such that:

f (f (n)) = = N

Where n is a bit signed integer; can ' t use complex numbers a Rithmetic.

If you can ' t-design such a function for the whole range of numbers, it is for the largest range possible.

Try one:

For languages that support the built-in state of a function, such as the C language, it is easy to solve this face test. The following code can be:

C + +

int f (int n) {

static int state =-1;

State *=-1;

return n * state;

}

This defines a function built in state variable states. The first time the function f is executed, State equals 1, and the function returns N. The second time the function is executed, state equals-1, and the function returns-N. This solution can be applied to all 32-bit integers except int_min. For int_min, negation causes overflow, i.e. F (f (int_min)) = Int_min.

Of course, this topic should not be so easy to solve. How do you solve a language that doesn't support built-in state, such as C # or Java?

Try two:

You can use global variables to store the state. For example, the following C # code:

private static int state =-1;

public static int f (int i) {State

*=-1;

return state * i;

}

But this solution is still not perfect, it is not so.