Rubyで2分探索

contains.rb

def contains(v, vs)
	if vs.length == 0
		return false;
	end
	left = 0; right = vs.length
	while left + 1 < right
		mid = left + (right - left) / 2;
		if v < vs[mid]
			right = mid
		else
			left = mid
		end
	end
	return v == vs[left];
end
 
vs = [*1..10000000]
p contains(100, vs)