Created
September 22, 2016 23:02
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Code example translated from Java to C
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/* Code example translated from Java to C */ | |
#include <stddef.h> | |
int minInt(int a, int b) | |
{ | |
return a < b ? a : b; | |
} | |
int maxInt(int a, int b) | |
{ | |
return a > b ? a : b; | |
} | |
int highestProductOf3(const int* arrayOfInts, size_t size) | |
{ | |
size_t i; | |
if (size < 3) | |
return 0; /* indicate error */ | |
/* We're going to start at the 3rd item (at index 2) | |
so pre-populate highests and lowests based on the first 2 items. | |
we could also start these as null and check below if they're set | |
but this is arguably cleaner */ | |
int highest = maxInt(arrayOfInts[0], arrayOfInts[1]); | |
int lowest = minInt(arrayOfInts[0], arrayOfInts[1]); | |
int highestProductOf2 = arrayOfInts[0] * arrayOfInts[1]; | |
int lowestProductOf2 = arrayOfInts[0] * arrayOfInts[1]; | |
/* except this one--we pre-populate it for the first /3/ items. | |
this means in our first pass it'll check against itself, which is fine. */ | |
int highestProductOf3 = arrayOfInts[0] * arrayOfInts[1] * arrayOfInts[2]; | |
/* walk through items, starting at index 2 */ | |
for (i = 2; i < size; i++) { | |
int current = arrayOfInts[i]; | |
/* do we have a new highest product of 3? | |
it's either the current highest, | |
or the current times the highest product of two | |
or the current times the lowest product of two */ | |
highestProductOf3 = maxInt(maxInt( | |
highestProductOf3, | |
current * highestProductOf2), | |
current * lowestProductOf2); | |
/* do we have a new highest product of two? */ | |
highestProductOf2 = maxInt(maxInt( | |
highestProductOf2, | |
current * highest), | |
current * lowest); | |
/* do we have a new lowest product of two? */ | |
lowestProductOf2 = minInt(minInt( | |
lowestProductOf2, | |
current * highest), | |
current * lowest); | |
/* do we have a new highest? */ | |
highest = maxInt(highest, current); | |
/* do we have a new lowest? */ | |
lowest = minInt(lowest, current); | |
} | |
return highestProductOf3; | |
} |
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