h*9
3 楼
I would do as below.
61 buckets -- one for each minute. last bucket is for the current minute.
every hit will increase the counter in current bucket by 1. at the end of
the minute, phase out the oldest bucket and create a new bucket as the
current bucket.
最后一分钟点击量 = bucket[60] + interpolate(bucket[59])
最后一小时的点击量 = interpolate(bucket[0]) + sum(bucket[1] ... bucket[60])
61 buckets -- one for each minute. last bucket is for the current minute.
every hit will increase the counter in current bucket by 1. at the end of
the minute, phase out the oldest bucket and create a new bucket as the
current bucket.
最后一分钟点击量 = bucket[60] + interpolate(bucket[59])
最后一小时的点击量 = interpolate(bucket[0]) + sum(bucket[1] ... bucket[60])
C*U
6 楼
我开始是这么和他说的
但是他后来又有写问题
说如果是1分零1秒怎么办
然后我就改到tree结构了
【在 h**********9 的大作中提到】
: I would do as below.
: 61 buckets -- one for each minute. last bucket is for the current minute.
: every hit will increase the counter in current bucket by 1. at the end of
: the minute, phase out the oldest bucket and create a new bucket as the
: current bucket.
: 最后一分钟点击量 = bucket[60] + interpolate(bucket[59])
: 最后一小时的点击量 = interpolate(bucket[0]) + sum(bucket[1] ... bucket[60])
但是他后来又有写问题
说如果是1分零1秒怎么办
然后我就改到tree结构了
【在 h**********9 的大作中提到】
: I would do as below.
: 61 buckets -- one for each minute. last bucket is for the current minute.
: every hit will increase the counter in current bucket by 1. at the end of
: the minute, phase out the oldest bucket and create a new bucket as the
: current bucket.
: 最后一分钟点击量 = bucket[60] + interpolate(bucket[59])
: 最后一小时的点击量 = interpolate(bucket[0]) + sum(bucket[1] ... bucket[60])
p*2
10 楼
把精确度先搞定,然后用滚动数组+counter就可以了。
C*U
13 楼
那就和我的那个树结构是一个道理。
e*s
15 楼
可以用24×60×60 Queue吗?
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