It's an O(n) algorithm.

Now we iterate both lists simultaneously to find a hole.

[xemul: Discussion making the patch more understandable:

Cyrill:

	If s_vma is the last one on self_vma_list you could break immediately, no?

	And the snippet I somehow miss is -- how the situation handled when

		      hole
		    a      b
	source |----|      |-----|
	target   |----|      |-----|
		      c      d

	the hole fits the requested size but the hole is shifted
	in target, so that you've

	prev_vma_end = a

	and then you find that a - d > vma_len and return a
	as start address for new mapping while finally it
	might intersect with address c.

	Or I miss something obvious?

Andrey:

	Look at "continue" one more time.
	prev_vma_end is returned only if both condition are true

	if (prev_vma_end + vma_len > s_vma->vma.start) {
	....
	if (prev_vma_end + vma_len > t_vma->vma.start) {
	...
Signed-off-by: Andrey Vagin <avagin@openvz.org>
Looks-good-to: Cyrill Gorcunov <gorcunov@openvz.org>
Signed-off-by: Pavel Emelyanov <xemul@parallels.com>