M*n
2 楼
申请绿卡的出生证公证是县公证处出具的,可以吗?
m*j
3 楼
bless
r*d
4 楼
you can get it wherever you can.
h*6
7 楼
如何用heap来实现呢,没有思路啊。
I*A
20 楼
how about this one?
maintain two heaps, max heap and min heap
for each number v
we compare it against max heap's max and min heap's min.
then, based on the comparison and the difference between the number of items
in min heap and max heap. we either
(1) put v into max or min directly or
(2) we might need to move max's max to min heap and push this v to max, or
(3) we might need to move min's min to max heap and push this v to min.
【在 l******4 的大作中提到】
: 还真是不对。 我这方法错了。
maintain two heaps, max heap and min heap
for each number v
we compare it against max heap's max and min heap's min.
then, based on the comparison and the difference between the number of items
in min heap and max heap. we either
(1) put v into max or min directly or
(2) we might need to move max's max to min heap and push this v to max, or
(3) we might need to move min's min to max heap and push this v to min.
【在 l******4 的大作中提到】
: 还真是不对。 我这方法错了。
l*4
21 楼
没看懂。
其实只要保证从小到大排列时max heap永远包含前半段, min heap永远包含后半段救
行。
那么还是我原来的方法,每插入一次min heap,检查一下min root 是否大于 max
root。 如
果不是,则交换roots。
one
the
items
【在 I**A 的大作中提到】
: how about this one?
: maintain two heaps, max heap and min heap
: for each number v
: we compare it against max heap's max and min heap's min.
: then, based on the comparison and the difference between the number of items
: in min heap and max heap. we either
: (1) put v into max or min directly or
: (2) we might need to move max's max to min heap and push this v to max, or
: (3) we might need to move min's min to max heap and push this v to min.
:
其实只要保证从小到大排列时max heap永远包含前半段, min heap永远包含后半段救
行。
那么还是我原来的方法,每插入一次min heap,检查一下min root 是否大于 max
root。 如
果不是,则交换roots。
one
the
items
【在 I**A 的大作中提到】
: how about this one?
: maintain two heaps, max heap and min heap
: for each number v
: we compare it against max heap's max and min heap's min.
: then, based on the comparison and the difference between the number of items
: in min heap and max heap. we either
: (1) put v into max or min directly or
: (2) we might need to move max's max to min heap and push this v to max, or
: (3) we might need to move min's min to max heap and push this v to min.
:
p*n
23 楼
You can find the problem in "Career cup 150".
Here is the question:
17.3 Numbers are randomly generated and stored in an array. Write a
program to
find and maintain the median value as new values are generated. pg 52
Solution #2: Keep an additional data structure (a tree)
Use two priority heaps: a max heap for the values below the median, and a
min heap for the values above the median. The median will be largest value
of the min heap. When a new value arrives it is placed in the below heap
if
Here is the question:
17.3 Numbers are randomly generated and stored in an array. Write a
program to
find and maintain the median value as new values are generated. pg 52
Solution #2: Keep an additional data structure (a tree)
Use two priority heaps: a max heap for the values below the median, and a
min heap for the values above the median. The median will be largest value
of the min heap. When a new value arrives it is placed in the below heap
if
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