speex: Fix resampler output length for interleaved input.

The original code has an assumption that if the resampler consumes
N input samples to produce M output samples, it assumes N samples is
enough to produce the M output samples. It is incorrect because the
resampler may actually used N+1 samples to produce the output, but it
only consumes N samples because the last input sample is still needed
(again) in the future.

The issue happens when the left channel resampler consumes N input
samples but actually used N+1, then the right channel resampler is only
given N input samples and cannot produce M samples.
The issue and fix is the same as mentioned in
http://lists.xiph.org/pipermail/speex-dev/2009-August/007406.html

--- speex-1.2rc1.orig/libspeex/resample.c	2008-05-30 15:34:16.000000000 +0800
+++ speex-1.2rc1/libspeex/resample.c	2013-09-13 20:24:07.981599107 +0800
@@ -954,13 +954,15 @@
 {
    spx_uint32_t i;
    int istride_save, ostride_save;
-   spx_uint32_t bak_len = *out_len;
+   spx_uint32_t bak_out_len = *out_len;
+   spx_uint32_t bak_in_len = *in_len;
    istride_save = st->in_stride;
    ostride_save = st->out_stride;
    st->in_stride = st->out_stride = st->nb_channels;
    for (i=0;i<st->nb_channels;i++)
    {
-      *out_len = bak_len;
+      *out_len = bak_out_len;
+      *in_len = bak_in_len;
       if (in != NULL)
          speex_resampler_process_float(st, i, in+i, in_len, out+i, out_len);
       else
@@ -975,13 +977,15 @@
 {
    spx_uint32_t i;
    int istride_save, ostride_save;
-   spx_uint32_t bak_len = *out_len;
+   spx_uint32_t bak_out_len = *out_len;
+   spx_uint32_t bak_in_len = *in_len;
    istride_save = st->in_stride;
    ostride_save = st->out_stride;
    st->in_stride = st->out_stride = st->nb_channels;
    for (i=0;i<st->nb_channels;i++)
    {
-      *out_len = bak_len;
+      *out_len = bak_out_len;
+      *in_len = bak_in_len;
       if (in != NULL)
          speex_resampler_process_int(st, i, in+i, in_len, out+i, out_len);
       else