$f(n)$ is : $Z^+\rightarrow Z$
$0$ is neither positive nor negative. Hence, $0 \notin Z^+$
$f(1) = 0; \hspace{0.5cm} f(3) = 1;\hspace{0.5cm} f(5) = 2;$ and so on…
$f(2) = -1;\hspace{0.5cm} f(4) = -2;\hspace{0.5cm} f(6) = -3;$ and so on…
We can see, $f(n)$ is One to One and Onto $\implies$ bijection.