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)