Poincare duality with boundary
Theorem. Let $M$ be an $n$-dimensional compact, oriented manifold with boundary, and $[M]\in H_n(M,\partial M)$ be its natural orientation class. Then
$$
H^i(M,\partial M)\overset{\cdot\cap[M]}{\longrightarrow} H_{n-i}(M)
$$
and
$$
H^i(M)\overset{\cdot\cap[M]}{\longrightarrow} H_{n-i}(M,\partial M)
$$
are both isomorphisms.
Alexander duality
Corollary. (Alexander duality in $S^n$) Let $K\subset S^n$ be a subcomplex of $S^n$ under some finite cell structure. Then there is a canonical isomorphism
$$
\tilde{H}^q(K)\cong\tilde{H}_{n-q-1}(S^n-K)
$$
Proof. Suppose that $N(K)$ is a regular neighborhood of $K$. Then $\forall q>0$, using Poincare duality with boundary:
$$
\begin{aligned}
\tilde{H}^q(K) &\cong H^q(K)\\
&\cong H^q(N(K))\\
&\cong H_{n-q}(N(K),\partial N(K))\\
&\cong H_{n-q}(S^n,S^n-K)\quad\\
&\cong H_{n-q-1}(S^n-K) =\tilde{H}_{n-q-1}(S^n-K).
\end{aligned}
$$
The last equality is because of the long exact sequence of homology induced by the pair $(S^n,S^n-K)$, and for $q\neq 0$, we have $H_{n-q}(S^n)=0$. Therefore, the case for $q\neq 0$ has been established. For $q=0$, notice that
$$
H_{n-1}(S^n-K)\oplus\mathbb{Z}\cong H_n(S^n,S^n-K)
$$
and
$$
\tilde{H}^0(K)\oplus\mathbb{Z}\cong H^0(K)
$$
Still using the above result: $H^0(K)\cong H_n(S^n,S^n-K)$, we have
$$
\tilde{H}_{n-1}(S^n-K)\cong \tilde{H}^0(K).
$$
This completes the case for $q=0$. $\square$
Knot complement
Using Alexander duality, the computation of homology groups of knot complement in $S^3$ is simple.
e.g. Suppose that $K\subset \mathbb{R}^3$ is a knot. Then by Alexander duality in $S^3$,
$$
H_1(S^3-K)\cong H^1(K)\cong H^1(S^1)\cong\mathbb{Z}
$$
and
$$
H_2(S^3-K)\cong \tilde{H}^0(K)=0.
$$