Can a capacitor be charged without having resistance in the circuit? I was looking for the fact when a capacitor is directly connected to battery without resistor what will happen? 
If possible then tell the time for charging.
 A: In the context of ideal circuit theory, if an ideal constant voltage source with voltage across $v_S = V_{DC}$ is, at time $t = 0$, instantaneously connected to an ideal, uncharged capacitor, the voltage across the capacitor is a step
$$v_C(t) = V_{DC}u(t)$$
and so the current through is an impulse
$$i_C(t) = CV_{DC}\delta(t)$$
This is clearly unphysical so there's something missing from the model. As others have pointed out, a physical voltage source cannot supply arbitrarily large current and so the voltage across the capacitor cannot instantaneously change (since the current through is finite, the voltage rate of change is finite).
In addition, the area enclosed by the source, conductors, and capacitor is not zero and so there is a self-inductance of the circuit and resistance of the conductors that can limit the instantaneous current and its rate of change.
Further, physical capacitors actually have an associated inductance and series resistance.
So, to properly model this using ideal circuit elements, all of these 'parasitic' inductances and resistances must be added to the ideal circuit model to more accurately predict the physical charging current.

From the comments:

The voltage at a capacitor can not "jump", this is also well known
  from circuit theory

In ideal circuit theory, the voltage across a capacitor can be discontinuous if the current through is an impulse.  As an example, and because of this push back from the comments, I'll post this screenshot from the book "Electric Circuits and Networks" (via Google books):

A: Every battery has an internal resistance. This the time of charging would be defined by the value of this resistance plus the resistance of the connecting cables, and finally by the internal resistance of the capacitor. In an ideal case of a superconducting battery and capacitor, the charging time would be defined by the inductive resistance of the connecting cables.
A: In the real world, each of the simple passive components (resistor, inductor, capacitor) contain a little of each other. That is, a resistor has an inductance, a capacitor has a resistance etc.
No matter how you try to minimise these effects, some will always remain. Your capacitor in the question will have its own small internal resistance, and also the battery or power supply that you use to charge the capacitor will also have its own resistance. The wires that you use to connect the capacitor to the supply will in turn have their own resistance.
These are important effects to take into account when you try and ask what happens in an extreme case, such as in your question.
A: Ideally, a capacitor is made of two plates separated by an isolator. Consequently, ideally there is an open circuit there.
If you connect the capacitor to a battery, as no current can flow, each plate would ideally inmediately acquire the same potential as the battery. You know that conductors ideally adquire the same potential all along them (in electrostatics).
However, as other answers say, there's always a resistive effect on wires and elements, and you will always have no instantanous load, but an exponential RC one.
A: Supposed, 
"I was looking for the fact when a capacitor is directly connected to battery without resistor what will happen?" 
means the theoretical case "... a capacitor not having the battery voltage (e.g. an unloaded one) is directly connected to a battery without impedance...", this case is the generalized case of 
Capacitor discharging through no load?,
where the battery has simply 0 voltage resulting in a short, since an ideal battery has no (inner) impedance.
In this case here we have the same contradiction at the exact time of switching/connecting, except that u2 is the battery voltage.
The  contradiction is again u1 <> u2. So the generalized equivalency is to define a number n1=n2 and at the same time n1<>n2.
This is why in reality these circuits can not exist.
It is a contradiction on the pure theoretical level. The statement in another answer
" In the context of ideal circuit theory, if an ideal constant voltage source with voltage ... across is, at time ...
, instantaneously connected to an ideal, uncharged capacitor, the voltage across the capacitor is a step
and so the current through is an impulse."
may be misleading, since a capacitor is also an ideal voltage supply in the exact time of connection. Or with an unloaded ideal capacitor, the ideal voltage source with zero impedance is connected to the unloaded ideal capacitor having also zero impedance resulting in a undefined contradiction, since it's an ideal shortcut (without any inductivities/resistors/capacitors involved) to an ideal voltage source. So v_s and v_c are not at all known, are not defined, can not be calculated in the very first moment of connection and it's more than doubtful that a step function can be calculated like stated in that answer. It's like connecting 2 ideal voltage sources with different voltages. So once again, there is no need ( if it's not even misleading) to argue with real circuits and it's unavoidable impedances, the circuit is already theoretically impossible resp. based on a contradiction. 
The last paragraph in the cited answer is again misleading: "So, to properly model this using ideal circuit elements, all of these 'parasitic' inductances and resistances must be added to the ideal circuit model to more accurately predict the physical charging current.", since "to more accurately predict the...current" should read "to avoid an unsolvable contradiction".
