Remove byte order logic from half-transparency lookups

[StephenCWills]

I think I see two issues. First, here is where we draw the half-transparent rectangle.

devilutionX/Source/engine.cpp

Lines 54 to 59 in 6648f08

	const uint32_t v = *pix;
	#if SDL_BYTEORDER == SDL_LIL_ENDIAN
	*pix = lookupTable[v & 0xFFFF] | (lookupTable[(v >> 16) & 0xFFFF] << 16);
	#else
	*pix = lookupTable[(v >> 16) & 0xFFFF] | (lookupTable[v & 0xFFFF] << 16);
	#endif

When we make use of this lookup table on big-endian systems, we swap the 16-bit values we pass into the lookup, but the 16-bit values themselves are in the system's native endian order. However, since we are interpreting the original pixel as a native-endian 32-bit value and also assigning *pix as a native-endian 32-bit value, I think it would be easiest to use native-endian order throughout. The little-endian logic doesn't swap any bytes so we can just remove the special case for big-endian systems.

And that just leaves us with making sure that we build the lookup table using native-endian order...

devilutionX/Source/engine/palette.cpp

Lines 110 to 115 in 6648f08

	#if SDL_BYTEORDER == SDL_LIL_ENDIAN
	const std::uint16_t index = i | (j << 8);
	#else
	const std::uint16_t index = j | (i << 8);
	#endif
	paletteTransparencyLookupBlack16[index] = paletteTransparencyLookup[0][i] | (paletteTransparencyLookup[0][j] << 8);

On big-endian systems, i is placed before j when computing index, but paletteTransparencyLookup[0][j] is placed before paletteTransparencyLookup[0][i] regardless of the endian order. This swaps the bytes on big-endian systems when you use this lookup table. Since we want to use native-endian order, we don't need to swap any bytes on any systems. Therefore, once again, we should be able to remove the special case for big-endian systems since we currently don't swap the bytes on little-endian systems.

This resolves #6871 (I assume)