H*m
2 楼
http://www.youtube.com/watch?v=kMMHOsYxqvk
太牛了,应该比我的tablet强太多了
尤其是对力度的dynamic range
太牛了,应该比我的tablet强太多了
尤其是对力度的dynamic range
s*m
3 楼
美国佐治亚州3岁女孩Alida患有一种罕见神经性疾病,必须靠氧气罐和呼吸管才能维持
生命。但幼小的她根本背不动这么沉重的氧气罐,于是Alida的父母从犹他州找来了治
疗犬Gibbs先生。从此不论Alida去哪,Gibbs先生都背着氧气瓶不离左右
生命。但幼小的她根本背不动这么沉重的氧气罐,于是Alida的父母从犹他州找来了治
疗犬Gibbs先生。从此不论Alida去哪,Gibbs先生都背着氧气瓶不离左右
w*f
4 楼
min-heap with only 5 element,
or top-k selection?
or top-k selection?
M*s
7 楼
还有根据孩子身上的气味变化提醒孩子准时服药的狗狗,都很有爱。
f*n
10 楼
Find k smallest numbers in a list of size n:
* Use max-heap of size k, iterate through list
O(n log k)
* Turn entire list into min-heap, then pop k times
O(n + k log n)
* Use worst-case linear-time selection algorithm to find the value of the k'
th smallest element, then partition the list with this value, and take the
first k elements
O(n)
* Use max-heap of size k, iterate through list
O(n log k)
* Turn entire list into min-heap, then pop k times
O(n + k log n)
* Use worst-case linear-time selection algorithm to find the value of the k'
th smallest element, then partition the list with this value, and take the
first k elements
O(n)
h*e
12 楼
It is only asking for the five smallest numbers. Forget about min-heap and
linear time selection algorithm. The constant factor is so small that by
just keeping five ints and do comparisons directly you would probably yield
a faster program than all the algorithms you listed here.
k'
【在 f*******n 的大作中提到】
: Find k smallest numbers in a list of size n:
: * Use max-heap of size k, iterate through list
: O(n log k)
: * Turn entire list into min-heap, then pop k times
: O(n + k log n)
: * Use worst-case linear-time selection algorithm to find the value of the k'
: th smallest element, then partition the list with this value, and take the
: first k elements
: O(n)
linear time selection algorithm. The constant factor is so small that by
just keeping five ints and do comparisons directly you would probably yield
a faster program than all the algorithms you listed here.
k'
【在 f*******n 的大作中提到】
: Find k smallest numbers in a list of size n:
: * Use max-heap of size k, iterate through list
: O(n log k)
: * Turn entire list into min-heap, then pop k times
: O(n + k log n)
: * Use worst-case linear-time selection algorithm to find the value of the k'
: th smallest element, then partition the list with this value, and take the
: first k elements
: O(n)
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