In the worst case, the freq_table of policy data is not sorted and
contains duplicate frequencies, this means that it needs to iterate
through the entire freq_table of policy to ensure each frequency is
unique in the freq_table of stats data, this has a time complexity of
O(N^2), where N is the number of frequencies in the freq_table of
policy.
However, if the policy.freq_table is already sorted and contains no
duplicate frequencies, it can reduce the time complexity of creating
stats.freq_table to O(N), the 'freq_table_sorted' field of policy data
can be used to indicate whether the policy.freq_table is sorted.
Signed-off-by: Liao Chang <liaochang1@huawei.com>
Acked-by: Viresh Kumar <viresh.kumar@linaro.org>
Reviewed-by: Dhruva Gole <d-gole@ti.com>
[ rjw: Fix typo in changelog, remove redundant parens ]
Signed-off-by: Rafael J. Wysocki <rafael.j.wysocki@intel.com>
/* Find valid-unique entries */
cpufreq_for_each_valid_entry(pos, policy->freq_table)
- if (freq_table_get_index(stats, pos->frequency) == -1)
+ if (policy->freq_table_sorted != CPUFREQ_TABLE_UNSORTED ||
+ freq_table_get_index(stats, pos->frequency) == -1)
stats->freq_table[i++] = pos->frequency;
stats->state_num = i;