Does there exist a function $f:[0,1]\to [0,1]\setminus\left\{\frac 1n, n\in \mathbb{N}^\*\right\}$ so that $f$ is 1-1, $f(0)=0$ and $f$ continuous at $0$? Does it have a closed form?

I know that as $[0,1], [0,1]\setminus\left\{\frac 1n, n\in \mathbb{N}^\*\right\}$ have the same cardinality $\mathfrak{c}$, at least one bijection between them exists. By swaping $0,x$ ($x$ is the unique number for which $f(x)=0$) I can obtain $f(0)=0$. But what about continuity at $0$?