Consider a sequence $x[n]$ of length N, and assume $X[k]$ is its N-point DFT given by $$X[k] = \sum_{n=0}^{N-1}{x[n]e^{-j\frac{2\pi}{N}nk}}$$ $k = 0, 1,..., N-1$, which is computed through an N-point FFT. If you wish to compute the even-indexed ($k=0,2,4,...$), or the odd-indexed ($k=1,3,5,...$) indexed samples of $X[k]$, you can proceed with the following:
Denote the even indexed samples of $X[k]$ as $X_e[k]$ : \begin{align} X_e[k] = X[2k] &= \sum_{n=0}^{N-1}{x[n]e^{-j\frac{2\pi}{N}n(2k)}} &\scriptstyle{\text{term 2k computes even samples of X[k]}}\\ X_e[k] &= \sum_{n=0}^{N-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} &\scriptstyle{\text{factor 2 is moved into N/2 term}}\\ X_e[k] &= \sum_{n=0}^{\frac{N}{2}-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} + \sum_{n=\frac{N}{2}}^{N-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} &\scriptstyle{\text{Divide the sum into 2 halves}}\\ X_e[k] &= \sum_{n=0}^{\frac{N}{2}-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} + \sum_{n=0}^{\frac{N}{2}-1}{x[n+\frac{N}{2}]e^{-j\frac{2\pi}{N/2}nk}} &\scriptstyle{\text{adjust the 2nd sum range}}\\ X_e[k] &= \sum_{n=0}^{\frac{N}{2}-1}{ (x[n]+x[n+\frac{N}{2}])e^{-j\frac{2\pi}{N/2}nk}} &\scriptstyle{\text{merge the 2 sums}}\\ \end{align}
Final form suggest that the required even samples of N-point DFT of the sequence $x[n]$ can be computed from an N/2-point DFT of a new sequence $x_{half}[n]= x[n]+x[n+ N/2]$ whose length is half that of $x[n]$, assuming N is even.
The signal $x_{half}[n]$ is simply computed by adding the first half of the signal $x[n]$ into its second half. (if the length N of signal x[n] is not even padd a zero to its tail to make it even)
The following matlab/octave code gives you the desired even indexed samples of X[k] without computing a full N-point DFT of x[n]:
Xe = fft ( x(1:N/2) + x(N/2 + 1 : end), N/2);
Note that because of the addition of the halves before the FFT, efficiency will degrade from a pure N/2-point FFT.
The case for the odd indexed samples proceeds similary but results in a more complex form which I would like to summarize in the following Matlab script:
% Let x[n] be the signal of length N
xc = x .* exp(-j*2*pi*[0:N-1]/N) ; % Multiply x[n] with a complex phase term
Xo = fft (xc(1:N/2) + xc(N/2 + 1, N), N/2); % odd indexed samples of X[k]
Due to a complex multiplication before the FFT, in this case of computing odd indices performance is further reduced but still preferable over a direct N-point FFT.
The following Matlab/Octave excerpt demonstrates the computation of both the even and the odd indexed samples.
N = 1024; % original sequence length
x = randn(1,N);
X = fft(x,N); % original DFT of X
Xe = fft( x(1:N/2) + x(N/2 + 1: N) , N/2); % Even samples only from N/2 point FFT
Xo = fft( ( x(1:N/2) - x(N/2+1:N) ).*exp(-j*pi*[0:2:N-2]/N) , N/2); % Odd samples only from N/2 point FFT
figure,stem( abs ( Xe - X(1:2:N)) ); title('Xe - X(1:2:N)');
figure,stem( abs ( Xo - X(2:2:N)) ); title('Xo - X(2:2:N)');
Due to numerical roundoff effects, the difference plotted by the stems will be nonzero but very small...