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As Fernando Muro points out in the comments, Sylvester's law of inertia is probably the easiest way to determine the signature. You diagonalize the symmetric matrix by the Gram-Schmidt process. This is essentially as easy as Gaussian elimination.

A symmetric matrix is equivalent to a bilinear form, with basis $\{v_1, \ldots, v_n\}$, and matrix entries inner products $\langle v_i,v_j\rangle, i,j\leq n$. If $\langle v_1,v_1\rangle \neq 0$, then replace $v_i$ with $v_i-v_1 \langle v_1,v_i\rangle/\langle v_1,v_1\rangle$, and proceed by induction (this has the effect of performing the same row and column operations, so as to keep the matrix symmetric). The only variation on the standard Gram-Schmidt is what to do when $\langle v_1,v_1\rangle=0$, which clearly means the form is indefinite. If $\langle v_1, v_i\rangle=0$ for all $i\leq n$, then you can proceed by induction. Otherwise, by permuting indices, you may assume that $\langle v_1,v_2\rangle \neq 0$. Replace $v_1$ with $v_2+tv_1$, such that $\langle v_2,v_2\rangle+2t\langle v_1,v_2\rangle\neq 0$, and proceed as with the usual Gram-Schmidt. Once you have diagonalized in this way, Sylvester's law implies that the signature of the diagonal matrix is the same as that of the original one.

Note added: Since you claim you're interested in computing the signatures of 4-manifolds, there's some special points one might be able to take advantage of. Certainly then the form will be integral unimodular, so presumably your symmetric matrix has integer entries and determinant $\pm 1$. If the 4-manifolds are smooth, then by Donaldson's theorem, if the intersection form is (positive) definite, then it is integrally equivalent to the diagonal form $1\oplus 1\oplus \cdots\oplus 1=1^n$. Otherwise, if the form is indefinite, then it is either odd and integrally equivalent to a diagonal form $1^k\oplus (-1)^{n-k}$, or it is even and there is a unique equivalence class for a given $n$ and signature. One can then perform the diagonalization process over $\mathbb{Z}$ using some version of the Euclidean algorithm, except in the even case (in which case there is a standard tridiagonal integral form). Actually, in the even case one could take a sum $\oplus \pm 1$ to make the form odd, and then diagonalize over the integers (a similar trick taking a sum with a hyperbolic form makes things integrally diagonalizable in the general case if you are also considering non-smooth manifolds which might have non-diagonalizable forms). See Milnor-Husemoller or Conway-Sloane for more details.

To diagonalize an odd indefinite symmetric bilinear form $B$ over $\mathbb{Z}$, first solve the equation $B(v_1,v_1)=\pm 1$ (this is where you can use some kind of Euclidean algorithm), then do the orthogonalization process to get $v_2,\ldots, v_n$ such that $B(v_1,v_i)=0, i\geq 2$, and induct.

As Fernando Muro points out in the comments, Sylvester's law of inertia is probably the easiest way to determine the signature. You diagonalize the symmetric matrix by the Gram-Schmidt process. This is essentially as easy as Gaussian elimination.

A symmetric matrix is equivalent to a bilinear form, with basis $\{v_1, \ldots, v_n\}$, and matrix entries inner products $\langle v_i,v_j\rangle, i,j\leq n$. If $\langle v_1,v_1\rangle \neq 0$, then replace $v_i$ with $v_i-v_1 \langle v_1,v_i\rangle/\langle v_1,v_1\rangle$, and proceed by induction (this has the effect of performing the same row and column operations, so as to keep the matrix symmetric). The only variation on the standard Gram-Schmidt is what to do when $\langle v_1,v_1\rangle=0$, which clearly means the form is indefinite. If $\langle v_1, v_i\rangle=0$ for all $i\leq n$, then you can proceed by induction. Otherwise, by permuting indices, you may assume that $\langle v_1,v_2\rangle \neq 0$. Replace $v_1$ with $v_2+tv_1$, such that $\langle v_2,v_2\rangle+2t\langle v_1,v_2\rangle\neq 0$, and proceed as with the usual Gram-Schmidt. Once you have diagonalized in this way, Sylvester's law implies that the signature of the diagonal matrix is the same as that of the original one.

Note added: Since you claim you're interested in computing the signatures of 4-manifolds, there's some special points one might be able to take advantage of. Certainly then the form will be integral unimodular, so presumably your symmetric matrix has integer entries and determinant $\pm 1$. If the 4-manifolds are smooth, then by Donaldson's theorem, if the intersection form is (positive) definite, then it is integrally equivalent to the diagonal form $1\oplus 1\oplus \cdots\oplus 1=1^n$. Otherwise, if the form is indefinite, then it is either odd and integrally equivalent to a diagonal form $1^k\oplus (-1)^{n-k}$, or it is even and there is a unique equivalence class for a given $n$ and signature. One can then perform the diagonalization process over $\mathbb{Z}$ using some version of the Euclidean algorithm, except in the even case (in which case there is a standard tridiagonal integral form). Actually, in the even case one could take a sum $\oplus \pm 1$ to make the form odd, and then diagonalize over the integers (a similar trick taking a sum with a hyperbolic form makes things integrally diagonalizable in the general case if you are also considering non-smooth manifolds which might have non-diagonalizable forms). See Milnor-Husemoller or Conway-Sloane for more details.

As Fernando Muro points out in the comments, Sylvester's law of inertia is probably the easiest way to determine the signature. You diagonalize the symmetric matrix by the Gram-Schmidt process. This is essentially as easy as Gaussian elimination.

A symmetric matrix is equivalent to a bilinear form, with basis $\{v_1, \ldots, v_n\}$, and matrix entries inner products $\langle v_i,v_j\rangle, i,j\leq n$. If $\langle v_1,v_1\rangle \neq 0$, then replace $v_i$ with $v_i-v_1 \langle v_1,v_i\rangle/\langle v_1,v_1\rangle$, and proceed by induction (this has the effect of performing the same row and column operations, so as to keep the matrix symmetric). The only variation on the standard Gram-Schmidt is what to do when $\langle v_1,v_1\rangle=0$, which clearly means the form is indefinite. If $\langle v_1, v_i\rangle=0$ for all $i\leq n$, then you can proceed by induction. Otherwise, by permuting indices, you may assume that $\langle v_1,v_2\rangle \neq 0$. Replace $v_1$ with $v_2+tv_1$, such that $\langle v_2,v_2\rangle+2t\langle v_1,v_2\rangle\neq 0$, and proceed as with the usual Gram-Schmidt. Once you have diagonalized in this way, Sylvester's law implies that the signature of the diagonal matrix is the same as that of the original one.

As Fernando Muro points out in the comments, Sylvester's law of inertia is probably the easiest way to determine the signature. You diagonalize the symmetric matrix by the Gram-Schmidt process. This is essentially as easy as Gaussian elimination.

A symmetric matrix is equivalent to a bilinear form, with basis $\{v_1, \ldots, v_n\}$, and matrix entries inner products $\langle v_i,v_j\rangle, i,j\leq n$. If $\langle v_1,v_1\rangle \neq 0$, then replace $v_i$ with $v_i-v_1 \langle v_1,v_i\rangle/\langle v_1,v_1\rangle$, and proceed by induction (this has the effect of performing the same row and column operations, so as to keep the matrix symmetric). The only variation on the standard Gram-Schmidt is what to do when $\langle v_1,v_1\rangle=0$, which clearly means the form is indefinite. If $\langle v_1, v_i\rangle=0$ for all $i\leq n$, then you can proceed by induction. Otherwise, by permuting indices, you may assume that $\langle v_1,v_2\rangle \neq 0$. Replace $v_1$ with $v_2+tv_1$, such that $\langle v_2,v_2\rangle+2t\langle v_1,v_2\rangle\neq 0$, and proceed as with the usual Gram-Schmidt. Once you have diagonalized in this way, Sylvester's law implies that the signature of the diagonal matrix is the same as that of the original one.

Note added: Since you claim you're interested in computing the signatures of 4-manifolds, there's some special points one might be able to take advantage of. Certainly then the form will be integral unimodular, so presumably your symmetric matrix has integer entries and determinant $\pm 1$. If the 4-manifolds are smooth, then by Donaldson's theorem, if the intersection form is (positive) definite, then it is integrally equivalent to the diagonal form $1\oplus 1\oplus \cdots\oplus 1=1^n$. Otherwise, if the form is indefinite, then it is either odd and integrally equivalent to a diagonal form $1^k\oplus (-1)^{n-k}$, or it is even and there is a unique equivalence class for a given $n$ and signature. One can then perform the diagonalization process over $\mathbb{Z}$ using some version of the Euclidean algorithm, except in the even case (in which case there is a standard tridiagonal integral form). Actually, in the even case one could take a sum $\oplus \pm 1$ to make the form odd, and then diagonalize over the integers (a similar trick taking a sum with a hyperbolic form makes things integrally diagonalizable in the general case if you are also considering non-smooth manifolds which might have non-diagonalizable forms). See Milnor-Husemoller or Conway-Sloane for more details.

To diagonalize an odd indefinite symmetric bilinear form $B$ over $\mathbb{Z}$, first solve the equation $B(v_1,v_1)=\pm 1$ (this is where you can use some kind of Euclidean algorithm), then do the orthogonalization process to get $v_2,\ldots, v_n$ such that $B(v_1,v_i)=0, i\geq 2$, and induct.

As Fernando Muro points out in the comments, Sylvester's law of inertia is probably the easiest way to determine the signature. You diagonalize the symmetric matrix by orthogonal transformations, by the Gram-Schmidt process. This is essentially as easy as Gaussian elimination.

A symmetric matrix is equivalent to a bilinear form, with basis $\{v_1, \ldots, v_n\}$, and matrix entries inner products $\langle v_i,v_j\rangle, i,j\leq n$. If $\langle v_1,v_1\rangle \neq 0$, then replace $v_i$ with $v_i-v_1 \langle v_1,v_i\rangle/\langle v_1,v_1\rangle$, and proceed by induction (this has the effect of performing the same row and column operations, so as to keep the matrix symmetric). The only variation on the standard Gram-Schmidt is what to do when $\langle v_1,v_1\rangle=0$, which clearly means the form is indefinite. If $\langle v_1, v_i\rangle=0$ for all $i\leq n$, then you can proceed by induction. Otherwise, by permuting indices, you may assume that $\langle v_1,v_2\rangle \neq 0$. Replace $v_1$ with $v_2+tv_1$, such that $\langle v_2,v_2\rangle+2t\langle v_1,v_2\rangle\neq 0$, and proceed as with the usual Gram-Schmidt. Once you have diagonalized in this way, Sylvester's law implies that the signature of the diagonal matrix is the same as that of the original one.

As Fernando Muro points out in the comments, Sylvester's law of inertia is probably the easiest way to determine the signature. You diagonalize the symmetric matrix by the Gram-Schmidt process. This is essentially as easy as Gaussian elimination.

A symmetric matrix is equivalent to a bilinear form, with basis $\{v_1, \ldots, v_n\}$, and matrix entries inner products $\langle v_i,v_j\rangle, i,j\leq n$. If $\langle v_1,v_1\rangle \neq 0$, then replace $v_i$ with $v_i-v_1 \langle v_1,v_i\rangle/\langle v_1,v_1\rangle$, and proceed by induction (this has the effect of performing the same row and column operations, so as to keep the matrix symmetric). The only variation on the standard Gram-Schmidt is what to do when $\langle v_1,v_1\rangle=0$, which clearly means the form is indefinite. If $\langle v_1, v_i\rangle=0$ for all $i\leq n$, then you can proceed by induction. Otherwise, by permuting indices, you may assume that $\langle v_1,v_2\rangle \neq 0$. Replace $v_1$ with $v_2+tv_1$, such that $\langle v_2,v_2\rangle+2t\langle v_1,v_2\rangle\neq 0$, and proceed as with the usual Gram-Schmidt. Once you have diagonalized in this way, Sylvester's law implies that the signature of the diagonal matrix is the same as that of the original one.