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December 8, 2017 10:44
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Binary Search Algorithm
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/* We can only use binary search if the input array of numbers is already sorted! */ | |
let binarySearch = (target, numbers) => { | |
/* find an apearance of target in numbers */ | |
/* | |
** we have lowLimit and highLimit of our range(numbers) | |
** so starting point is -1 (the 0th index) | |
*/ | |
let lowLimit = -1; | |
let highLimit = numbers.length; | |
/* | |
** if there isn't at least 1 index between lowLimit and highLimit, | |
** we've run out of guesses and the number must not be present | |
*/ | |
while ( (lowLimit + 1) < highLimit ) { | |
/* | |
** find the index ~halfway between the lowLimit and highLimit | |
** we have to round down, to avoid getting a "half index" | |
*/ | |
let distance = highLimit - lowLimit; | |
let halfDistance = Math.floor(distance / 2); | |
let guessIndex = lowLimit + halfDistance; | |
let guessValue = numbers[guessIndex]; | |
if (guessValue === target) { | |
return true; | |
} | |
if (guessValue > target) { | |
/* target is to the left, so move high boundary to the left */ | |
highLimit = guessIndex; | |
} else { | |
/* target is to the right, so move low boundary to the right */ | |
lowLimit = guessIndex; | |
} | |
} | |
return false; | |
} |
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