问题
If every value one adds to a sorted set (redis) is one with the highest score, will the time complexity be O(log(N)) for each zadd?
OR, for such edge cases, redis performs optimizations (e.g. an exception that in such cases where score is higher than the highest score in the set, simply add the value at the highest spot)?
Practically, I ask because I keep a global sorted set in my app where values are zadded with time since epoch as the score. And I'm wondering whether this will still be O(log(N)), or would it be faster?
回答1:
Once a Sorted Set has grown over the thresholds set by the zset-max-ziplist-* configuration directives, it is encoded as a skip list. Optimizing insertion for this edge case seems impossible due to the need to maintain the skip list's upper levels. A cursory review of the source code shows that, as expected, this isn't handled in any special way.
来源：https://stackoverflow.com/questions/40870081/time-complexity-of-zadd-when-value-has-score-greater-than-highest-score-present