This fixes a corner case that is caused by a race of dio write vs dio
read/write.

Here is how the race could happen.

Suppose that no extent map has been loaded into memory yet.
There is a file extent [0, 32K), two jobs are running concurrently
against it, t1 is doing dio write to [8K, 32K) and t2 is doing dio
read from [0, 4K) or [4K, 8K).

t1 goes ahead of t2 and splits em [0, 32K) to em [0K, 8K) and [8K 32K).

------------------------------------------------------
             t1                                t2
      btrfs_get_blocks_direct()         btrfs_get_blocks_direct()
       -> btrfs_get_extent()              -> btrfs_get_extent()
           -> lookup_extent_mapping()
           -> add_extent_mapping()            -> lookup_extent_mapping()
              # load [0, 32K)
       -> btrfs_new_extent_direct()
           -> btrfs_drop_extent_cache()
              # split [0, 32K) and
	      # drop [8K, 32K)
           -> add_extent_mapping()
              # add [8K, 32K)
                                              -> add_extent_mapping()
                                                 # handle -EEXIST when adding
                                                 # [0, 32K)
------------------------------------------------------
About how t2(dio read/write) runs into -EEXIST:

a) add_extent_mapping() gets -EEXIST for adding em [0, 32k),

b) search_extent_mapping() then returns [0, 8k) as the existing em,
   even though start == existing->start, em is [0, 32k) so that
   extent_map_end(em) > extent_map_end(existing), i.e. 32k > 8k,

c) then it goes thru merge_extent_mapping() which tries to add a [8k, 8k)
   (with a length 0) and returns -EEXIST as [8k, 32k) is already in tree,

d) so btrfs_get_extent() ends up returning -EEXIST to dio read/write,
   which is confusing applications.

Here I conclude all the possible situations,
1) start < existing->start

            +-----------+em+-----------+
+--prev---+ |     +-------------+      |
|         | |     |             |      |
+---------+ +     +---+existing++      ++
                +
                |
                +
             start

2) start == existing->start

      +------------em------------+
      |     +-------------+      |
      |     |             |      |
      +     +----existing-+      +
            |
            |
            +
         start

3) start > existing->start && start < (existing->start + existing->len)

      +------------em------------+
      |     +-------------+      |
      |     |             |      |
      +     +----existing-+      +
               |
               |
               +
             start

4) start >= (existing->start + existing->len)

+-----------+em+-----------+
|     +-------------+      | +--next---+
|     |             |      | |         |
+     +---+existing++      + +---------+
                      +
                      |
                      +
                   start

As we can see, it turns out that if start is within existing em (front
inclusive), then the existing em should be returned as is, otherwise,
we try our best to merge candidate em with sibling ems to form a
larger em (in order to reduce the total number of em).
Reported-by: David Vallender <david.vallender@landmark.co.uk>
Signed-off-by: Liu Bo <bo.li.liu@oracle.com>
Reviewed-by: Josef Bacik <jbacik@fb.com>
Signed-off-by: David Sterba <dsterba@suse.com>