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臭臭丁三岁生日包,7:00到8:00有效
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臭臭丁三岁生日包,7:00到8:00有效# pets - 心有所宠
P*n
1 楼
【 以下文字转载自 FleaMarket 讨论区 】
发信人: Purpleseven (Purple7紫七), 信区: FleaMarket
标 题: ▄▇征护士▇▄ 长期大量收 Outlet Coach 长期合作 (转载)
发信站: BBS 未名空间站 (Tue May 4 15:45:59 2010, 美东)
发信人: Purpleseven (Purple7紫七), 信区: shopping
标 题: ▄▇征护士▇▄ 长期大量收 Outlet Coach 长期合作
发信站: BBS 未名空间站 (Fri Apr 30 10:35:51 2010, 美东)
▄▇征护士▇▄ 长期大量收 Outlet Coach 长期合作
（要联系 请用后面联系方式 不要给我发信 谢谢）
各位亲爱的护士们，大家好！~ 我是Purpleseven的助理，之后由我负责组织Coach及其他项目代
购收货。我们期望能够在这里找到合适长期负责收货的护士。
优先次序：
1． 免税州 或 低税 的护士优先 （有Gift card渠道的最好）
2。 有经验的，有好评的护士优先
3。 发货专业，之前有过合作的护
avatar
l*r
2 楼
Given the root of a binary tree (not balanced), heapify it.
Am I right we msut need the parent pointer for each node to be more
efficient?
avatar
a*n
3 楼
会不会被律师报复把我的买房手续胡搞？俩都中国人，倍儿诈的那种。
贷款可以换，可律师换不了啦。。。
avatar
t*g
5 楼
臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
啦啦
avatar
p*r
6 楼
免税州+coupon+gc...
这个能有dollar menu burger就不错了

其他项目代

【在 P*********n 的大作中提到】
: 【 以下文字转载自 FleaMarket 讨论区 】
: 发信人: Purpleseven (Purple7紫七), 信区: FleaMarket
: 标 题: ▄▇征护士▇▄ 长期大量收 Outlet Coach 长期合作 (转载)
: 发信站: BBS 未名空间站 (Tue May 4 15:45:59 2010, 美东)
: 发信人: Purpleseven (Purple7紫七), 信区: shopping
: 标 题: ▄▇征护士▇▄ 长期大量收 Outlet Coach 长期合作
: 发信站: BBS 未名空间站 (Fri Apr 30 10:35:51 2010, 美东)
: ▄▇征护士▇▄ 长期大量收 Outlet Coach 长期合作
: （要联系 请用后面联系方式 不要给我发信 谢谢）
: 各位亲爱的护士们，大家好！~ 我是Purpleseven的助理，之后由我负责组织Coach及其他项目代
avatar
l*r
7 楼
By the way, my algorithm is like (assume max heap):
for the current node, if left or right is bigger than the current, switch
the biggest value as the current value, and try to move it all the way up to
its parent chain if that is bigger than the parent's value as well.
Then do the same for its left and right children.
avatar
b*d
8 楼
交友8被公司当做色情网站屏蔽了
avatar
K*a
9 楼
生日快乐臭臭丁
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l*d
10 楼
不是很多人联系你吗？怎么还在招护士啊？
avatar
l*r
11 楼
Any comments?
avatar
u*3
13 楼
Happy Birthday!!

【在 K***a 的大作中提到】
: 生日快乐臭臭丁
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P*n
14 楼
不够用 这批收的比较多
avatar
s*n
15 楼
It's still not balanced by your algorithm?
avatar
M*e
16 楼
看名字

【在 G*********s 的大作中提到】
: 为啥实诚
: 莫非博导在打广告？
avatar
l*8
17 楼
Happy Birthday!
avatar
l*r
18 楼
Why not?
It starts with root, it guarantee the current node and its two children
satisfy the condition. Then it back check (heapify) all its parents, and
guarantee that. Then move down to its children. I just don't know whether it
is efficient or not.

【在 s*****n 的大作中提到】
: It's still not balanced by your algorithm?
avatar
l*o
19 楼
劲敌
avatar
c*n
20 楼
生日快乐。嘿嘿，早起的鸟有包子吃。

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
s*n
21 楼
左右子节点都是balanced并不保证current node是balanced
比如左深度3 右深度100
avatar
M*n
22 楼
博导嫉妒了
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y*u
23 楼
生日快乐
avatar
c*y
24 楼
Does Post-Order travel work here?
void heapify(binaryTree *node)
{
if (!node) return;
if (node->left)
heapify(node->left);
if (node->right)
heapify(node->right);
int max = node->value;
if (node->left && node->left->value > max) {
max = node->left->value;
}
if (node->right && node->right->value > max) {
max = mode->right->value;
}
if (max == node->value) {
return;
} else if (node->left && max == node->left->value) {
int tmp = node->value;
node->value = node->left->value;
node->left->value = tmp;
} else {
int tmp = node->value;
node->value = node->right->value;
node->right->value = tmp;
}
return;
}
avatar
d*k
25 楼
大家观察力都很强大

【在 M*****e 的大作中提到】
: 看名字
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I*s
26 楼
生日快乐！！
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l*r
27 楼
"Given the root of a binary tree (not balanced), heapify it."
It asks to heapify, it does not ask to balance the tree.

【在 s*****n 的大作中提到】
: 左右子节点都是balanced并不保证current node是balanced
: 比如左深度3 右深度100
avatar
I*s
28 楼
大家都好早哦！
困。。。
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s*x
29 楼

I think you need push all the way down in the last swap step, thinking the
case root is the smallest value.

【在 c**y 的大作中提到】
: Does Post-Order travel work here?
: void heapify(binaryTree *node)
: {
: if (!node) return;
: if (node->left)
: heapify(node->left);
: if (node->right)
: heapify(node->right);
: int max = node->value;
: if (node->left && node->left->value > max) {
avatar
m*u
30 楼
Happy Birthday!
可惜没包子吃了。
avatar
c*y
31 楼
Thanks for pointing out the issue. I now know it.

【在 s**x 的大作中提到】
:
: I think you need push all the way down in the last swap step, thinking the
: case root is the smallest value.
avatar
p*7
32 楼
Happy Birthday!
avatar
g*e
33 楼

but heap is a balanced tree

【在 l********r 的大作中提到】
: "Given the root of a binary tree (not balanced), heapify it."
: It asks to heapify, it does not ask to balance the tree.
avatar
B*A
34 楼
Happy birthday

★ 发自iPhone App: ChineseWeb 7.7

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
c*y
35 楼
Revised solution. Will try to test it if I get a chance today.
void heapify(binaryTree *node)
{
if (!node) return;
if (node->left)
heapify(node->left);
if (node->right)
heapify(node->right);
int max = node->value;
if (node->left && node->left->value > max) {
max = node->left->value;
}
if (node->right && node->right->value > max) {
max = mode->right->value;
}
if (max == node->value) {
return;
} else if (node->left && max == node->left->value) {
int tmp = node->value;
node->value = node->left->value;
node->left->value = tmp;
heapify(node->left);
} else {
int tmp = node->value;
node->value = node->right->value;
node->right->value = tmp;
heapify(node->right);
}
return;
}
avatar
l*o
36 楼
生日快乐
avatar
s*x
37 楼

the pushdown (or siftdown) part should not call heapify, it is much simpler
than heapify, just swap the bigger of two children with root if needed, you
need a separate function or a for loop I think.

【在 c**y 的大作中提到】
: Revised solution. Will try to test it if I get a chance today.
: void heapify(binaryTree *node)
: {
: if (!node) return;
: if (node->left)
: heapify(node->left);
: if (node->right)
: heapify(node->right);
: int max = node->value;
: if (node->left && node->left->value > max) {
avatar
o*d
38 楼
Happy b day!

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
c*y
39 楼
The revised algorithm above follows the approach used in building a heap,
which consists of two parts: BUILD-MAX-HEAP (see page 157 of CLRC, 3rd
edition) and MAX-HEAPIFY (see page 154 of the same book).
The MAX-HEAPIFY algorithm is a recursive solution in that it calls itself
again in the pushdown part. Similarly, heapify is directly called by itself
in the above solution.
BUILD-MAX-HEAP restores the heap property in a bottom-up fashion.
Correspondingly, post-order is used to achieve the same bottom-up effect.

simpler
you

【在 s**x 的大作中提到】
:
: the pushdown (or siftdown) part should not call heapify, it is much simpler
: than heapify, just swap the bigger of two children with root if needed, you
: need a separate function or a for loop I think.
avatar
z*i
40 楼
bless!

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
s*x
41 楼
I just reviewed some chapters and searched online as well. the MAX-HEAPITY
is called SiftDown in some other papers, which makes more sense.
MAX-HEAPITY is SiftDown
BUILD-MAX-HEAP is calling SiftDown/MAX-HEAPITY from bottom up.
your solution does not have a SiftDown Call, siftdown only touches one child
, while your heapify always touches both left and right children during a
recursive call.it might work too, but would be expensive.

itself

【在 c**y 的大作中提到】
: The revised algorithm above follows the approach used in building a heap,
: which consists of two parts: BUILD-MAX-HEAP (see page 157 of CLRC, 3rd
: edition) and MAX-HEAPIFY (see page 154 of the same book).
: The MAX-HEAPIFY algorithm is a recursive solution in that it calls itself
: again in the pushdown part. Similarly, heapify is directly called by itself
: in the above solution.
: BUILD-MAX-HEAP restores the heap property in a bottom-up fashion.
: Correspondingly, post-order is used to achieve the same bottom-up effect.
:
: simpler
avatar
d*g
42 楼
没包子也生日快乐~~臭臭丁要健健康康的!!
avatar
s*x
43 楼
modifies Caiy's code:
void SiftDown(binaryTree *node)
{
if(!node) return;

binaryTree *largest = node;
if (node->left != NULL && largest->value < node->left->value)
largest = node->left;
if (node->right != NULL && largest->value < node->right->value)
largest = node->right;
if (largest == node) return;

int tmp = node->value;
node->value = largest->value;
largest->value = tmp;
SiftDown(largest);
}
void heapify(binaryTree *node)
{
if (!node) return;
if (node->left)
heapify(node->left);
if (node->right)
heapify(node->right);
SiftDown(node);
}
avatar
c*a
44 楼
生日快乐！
你这么早蒸包子，让我们西岸的情何以堪啊！
avatar
l*r
45 楼
Given the root of a binary tree (not balanced), heapify it.
Am I right we msut need the parent pointer for each node to be more
efficient?
avatar
k*e
46 楼
生日快乐！
哪有生日包还限时1小时的！
avatar
l*r
47 楼
By the way, my algorithm is like (assume max heap):
for the current node, if left or right is bigger than the current, switch
the biggest value as the current value, and try to move it all the way up to
its parent chain if that is bigger than the parent's value as well.
Then do the same for its left and right children.
avatar
S*t
48 楼
好小啊 happy birthday

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
l*r
49 楼
Any comments?
avatar
M*i
50 楼
生日快乐

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
s*n
51 楼
It's still not balanced by your algorithm?
avatar
m*u
52 楼
没效了一样bless
avatar
l*r
53 楼
Why not?
It starts with root, it guarantee the current node and its two children
satisfy the condition. Then it back check (heapify) all its parents, and
guarantee that. Then move down to its children. I just don't know whether it
is efficient or not.

【在 s*****n 的大作中提到】
: It's still not balanced by your algorithm?
avatar
a*t
54 楼
Happy Birthday!
avatar
s*n
55 楼
左右子节点都是balanced并不保证current node是balanced
比如左深度3 右深度100
avatar
t*u
56 楼
生日快乐
你家还真是都够抠的呀
avatar
c*y
57 楼
Does Post-Order travel work here?
void heapify(binaryTree *node)
{
if (!node) return;
if (node->left)
heapify(node->left);
if (node->right)
heapify(node->right);
int max = node->value;
if (node->left && node->left->value > max) {
max = node->left->value;
}
if (node->right && node->right->value > max) {
max = mode->right->value;
}
if (max == node->value) {
return;
} else if (node->left && max == node->left->value) {
int tmp = node->value;
node->value = node->left->value;
node->left->value = tmp;
} else {
int tmp = node->value;
node->value = node->right->value;
node->right->value = tmp;
}
return;
}
avatar
m*f
58 楼
happy bday~~~
avatar
l*r
59 楼
"Given the root of a binary tree (not balanced), heapify it."
It asks to heapify, it does not ask to balance the tree.

【在 s*****n 的大作中提到】
: 左右子节点都是balanced并不保证current node是balanced
: 比如左深度3 右深度100
avatar
l*3
60 楼
happy b-day!
avatar
s*x
61 楼

I think you need push all the way down in the last swap step, thinking the
case root is the smallest value.

【在 c**y 的大作中提到】
: Does Post-Order travel work here?
: void heapify(binaryTree *node)
: {
: if (!node) return;
: if (node->left)
: heapify(node->left);
: if (node->right)
: heapify(node->right);
: int max = node->value;
: if (node->left && node->left->value > max) {
avatar
r*8
62 楼
生日快乐~~
才三岁，正是青壮年泡m的时候哈

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
c*y
63 楼
Thanks for pointing out the issue. I now know it.

【在 s**x 的大作中提到】
:
: I think you need push all the way down in the last swap step, thinking the
: case root is the smallest value.
avatar
Y*y
64 楼
哇哇哇~生日快乐~！！
avatar
g*e
65 楼

but heap is a balanced tree

【在 l********r 的大作中提到】
: "Given the root of a binary tree (not balanced), heapify it."
: It asks to heapify, it does not ask to balance the tree.
avatar
b*g
66 楼
生日快乐!!!
avatar
c*y
67 楼
Revised solution. Will try to test it if I get a chance today.
void heapify(binaryTree *node)
{
if (!node) return;
if (node->left)
heapify(node->left);
if (node->right)
heapify(node->right);
int max = node->value;
if (node->left && node->left->value > max) {
max = node->left->value;
}
if (node->right && node->right->value > max) {
max = mode->right->value;
}
if (max == node->value) {
return;
} else if (node->left && max == node->left->value) {
int tmp = node->value;
node->value = node->left->value;
node->left->value = tmp;
heapify(node->left);
} else {
int tmp = node->value;
node->value = node->right->value;
node->right->value = tmp;
heapify(node->right);
}
return;
}
avatar
b*b
68 楼
Happy Birthday!!
avatar
s*x
69 楼

the pushdown (or siftdown) part should not call heapify, it is much simpler
than heapify, just swap the bigger of two children with root if needed, you
need a separate function or a for loop I think.

【在 c**y 的大作中提到】
: Revised solution. Will try to test it if I get a chance today.
: void heapify(binaryTree *node)
: {
: if (!node) return;
: if (node->left)
: heapify(node->left);
: if (node->right)
: heapify(node->right);
: int max = node->value;
: if (node->left && node->left->value > max) {
avatar
t*g
70 楼
嘿嘿,好象早上8点前没什么人,大家那么热情就都给包子吧
avatar
c*y
71 楼
The revised algorithm above follows the approach used in building a heap,
which consists of two parts: BUILD-MAX-HEAP (see page 157 of CLRC, 3rd
edition) and MAX-HEAPIFY (see page 154 of the same book).
The MAX-HEAPIFY algorithm is a recursive solution in that it calls itself
again in the pushdown part. Similarly, heapify is directly called by itself
in the above solution.
BUILD-MAX-HEAP restores the heap property in a bottom-up fashion.
Correspondingly, post-order is used to achieve the same bottom-up effect.

simpler
you

【在 s**x 的大作中提到】
:
: the pushdown (or siftdown) part should not call heapify, it is much simpler
: than heapify, just swap the bigger of two children with root if needed, you
: need a separate function or a for loop I think.
avatar
m*s
72 楼
我算最后一个吧。

【在 t*****g 的大作中提到】
: 嘿嘿,好象早上8点前没什么人,大家那么热情就都给包子吧
avatar
s*x
73 楼
I just reviewed some chapters and searched online as well. the MAX-HEAPITY
is called SiftDown in some other papers, which makes more sense.
MAX-HEAPITY is SiftDown
BUILD-MAX-HEAP is calling SiftDown/MAX-HEAPITY from bottom up.
your solution does not have a SiftDown Call, siftdown only touches one child
, while your heapify always touches both left and right children during a
recursive call.it might work too, but would be expensive.

itself

【在 c**y 的大作中提到】
: The revised algorithm above follows the approach used in building a heap,
: which consists of two parts: BUILD-MAX-HEAP (see page 157 of CLRC, 3rd
: edition) and MAX-HEAPIFY (see page 154 of the same book).
: The MAX-HEAPIFY algorithm is a recursive solution in that it calls itself
: again in the pushdown part. Similarly, heapify is directly called by itself
: in the above solution.
: BUILD-MAX-HEAP restores the heap property in a bottom-up fashion.
: Correspondingly, post-order is used to achieve the same bottom-up effect.
:
: simpler
avatar
s*i
74 楼
bless

【在 t*****g 的大作中提到】
: 臭臭丁三岁了，发包子,不限个数,需要宠版ID,8点截止,就一个小时大家快来祝福吧!啦
: 啦啦
avatar
s*x
75 楼
modifies Caiy's code:
void SiftDown(binaryTree *node)
{
if(!node) return;

binaryTree *largest = node;
if (node->left != NULL && largest->value < node->left->value)
largest = node->left;
if (node->right != NULL && largest->value < node->right->value)
largest = node->right;
if (largest == node) return;

int tmp = node->value;
node->value = largest->value;
largest->value = tmp;
SiftDown(largest);
}
void heapify(binaryTree *node)
{
if (!node) return;
if (node->left)
heapify(node->left);
if (node->right)
heapify(node->right);
SiftDown(node);
}
avatar
y*a
76 楼
祝福还要看时效？
生日快乐
avatar
c*p
77 楼
mark
avatar
y*a
78 楼

haha,铁公鸡之家

【在 t*****u 的大作中提到】
: 生日快乐
: 你家还真是都够抠的呀
avatar
x*0
79 楼
mark
avatar
N*t
80 楼
生日快乐！一定要包子包子包子
avatar
c*p
81 楼
mark
avatar
x*0
82 楼
mark
avatar
x*a
83 楼
不太理解这里的heapify
比如说一个heap, 是一个complete binary tree. 所以其实他是balanced的.
如果不需要上面这个条件, 只是让他有heap的特征, 比如说root就是max或者min, post
order traversal就可以, 不需要parent pointer
avatar
s*x
84 楼
先读一读几个 heap sort 的算法就明白了。 不同的书用的术语也不一样， 核心是一
样的。 我感觉你取post order 也是一样的。不过没人那么说而已，本质都是先处理子
树，再处理根结点，处理根结点就是siftdown.

post
★ 发自iPhone App: ChineseWeb 8.6

【在 x*****a 的大作中提到】
: 不太理解这里的heapify
: 比如说一个heap, 是一个complete binary tree. 所以其实他是balanced的.
: 如果不需要上面这个条件, 只是让他有heap的特征, 比如说root就是max或者min, post
: order traversal就可以, 不需要parent pointer
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