convertible car seat上面没有遮太阳的# Parenting - 为人父母
b*z
1 楼
弱问一个以前听说有绿卡（刚毕业PhD或是薄厚）才能申请的NIH、NSF的Funding。忘了
名字？
名字？
c*n
2 楼
It is yesterday.
I caught very bad cold for about a week and it's close to the worst when I
interviewed with amazon. But I still went as arranged. For I have confidence
in my Lord.
I met 6 people. The first one is an HR staff. The rest 5 are engineers and
product managers. A product manager had lunch with me, which was also part
of the interview.
Omitting the first person's questions. (why amazon? Why physics --> software
?)
The second person asked me 2 easy questions: reverse string and find al
I caught very bad cold for about a week and it's close to the worst when I
interviewed with amazon. But I still went as arranged. For I have confidence
in my Lord.
I met 6 people. The first one is an HR staff. The rest 5 are engineers and
product managers. A product manager had lunch with me, which was also part
of the interview.
Omitting the first person's questions. (why amazon? Why physics --> software
?)
The second person asked me 2 easy questions: reverse string and find al
x*m
3 楼
宝宝朝后面坐着，有时太阳很大，直接晒到脸上，大家都单买遮太阳的吗？我没看到有
卖的啊
卖的啊
b*z
4 楼
顶一个。
没人知道么？
【在 b***z 的大作中提到】
: 弱问一个以前听说有绿卡（刚毕业PhD或是薄厚）才能申请的NIH、NSF的Funding。忘了
: 名字？
没人知道么？
【在 b***z 的大作中提到】
: 弱问一个以前听说有绿卡（刚毕业PhD或是薄厚）才能申请的NIH、NSF的Funding。忘了
: 名字？
f*5
5 楼
congratulations
could u say something about your phone interview?
thanks
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
could u say something about your phone interview?
thanks
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
a*n
6 楼
同问。
h*g
7 楼
NIH K series except K99, such as K25
f*5
8 楼
could u share your result for below?
But finally, I came back to the classical tree-algorithm. I
almost got the correct answer -- with a little piece missing.
when I
confidence
and
part
software
all the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
But finally, I came back to the classical tree-algorithm. I
almost got the correct answer -- with a little piece missing.
when I
confidence
and
part
software
all the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
x*t
9 楼
Get some sun shades for the windows. for example:
http://www.greatbabyproducts.com/dora-sun-block-car-roller-shad
http://www.greatbabyproducts.com/dora-sun-block-car-roller-shad
p*x
10 楼
K01, K23, K25都是
Career原来以讹传讹是，其实不是
【在 b***z 的大作中提到】
: 弱问一个以前听说有绿卡（刚毕业PhD或是薄厚）才能申请的NIH、NSF的Funding。忘了
: 名字？
Career原来以讹传讹是，其实不是
【在 b***z 的大作中提到】
: 弱问一个以前听说有绿卡（刚毕业PhD或是薄厚）才能申请的NIH、NSF的Funding。忘了
: 名字？
g*y
11 楼
恭喜物理转CS的同志！
欢饮来西雅图
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
欢饮来西雅图
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
y*i
12 楼
这东西对于后窗晒过来的太阳用处不大。
我家的carseat上自带的棚子可以转到放脚的那头，孩子朝后坐的时候可以用来遮阳。
【在 x******t 的大作中提到】
: Get some sun shades for the windows. for example:
: http://www.greatbabyproducts.com/dora-sun-block-car-roller-shad
我家的carseat上自带的棚子可以转到放脚的那头，孩子朝后坐的时候可以用来遮阳。
【在 x******t 的大作中提到】
: Get some sun shades for the windows. for example:
: http://www.greatbabyproducts.com/dora-sun-block-car-roller-shad
b*z
13 楼
谢过了。
【在 p**x 的大作中提到】
: K01, K23, K25都是
: Career原来以讹传讹是，其实不是
【在 p**x 的大作中提到】
: K01, K23, K25都是
: Career原来以讹传讹是，其实不是
s*l
14 楼
cong~
could you tell us how you designed for this problem?
"Then his questions was "design an architecture such that it has
best availability and scalability"
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
could you tell us how you designed for this problem?
"Then his questions was "design an architecture such that it has
best availability and scalability"
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
k*b
15 楼
戴顶遮阳帽?
c*n
16 楼
could u share your result for below?
But finally, I came back to the classical tree-algorithm. I
almost got the correct answer -- with a little piece missing.
basically the solution is very similar to in-order traversal: you just need
to count how many nodes you have traversed so far. As soon as you got enough
, you return that node. If you finished without reaching the number, you
return null. You need to define a pair structure as the return
value, where int represents the number o
But finally, I came back to the classical tree-algorithm. I
almost got the correct answer -- with a little piece missing.
basically the solution is very similar to in-order traversal: you just need
to count how many nodes you have traversed so far. As soon as you got enough
, you return that node. If you finished without reaching the number, you
return null. You need to define a pair
value, where int represents the number o
x*m
17 楼
我们只有infant car seat有棚子，换成convertible后貌似只有一个座椅，没有棚子了
【在 y****i 的大作中提到】
: 这东西对于后窗晒过来的太阳用处不大。
: 我家的carseat上自带的棚子可以转到放脚的那头，孩子朝后坐的时候可以用来遮阳。
【在 y****i 的大作中提到】
: 这东西对于后窗晒过来的太阳用处不大。
: 我家的carseat上自带的棚子可以转到放脚的那头，孩子朝后坐的时候可以用来遮阳。
b*v
18 楼
cong! thanks for sharing!
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
confidence
software
the
【在 c*********n 的大作中提到】
: It is yesterday.
: I caught very bad cold for about a week and it's close to the worst when I
: interviewed with amazon. But I still went as arranged. For I have confidence
: in my Lord.
: I met 6 people. The first one is an HR staff. The rest 5 are engineers and
: product managers. A product manager had lunch with me, which was also part
: of the interview.
: Omitting the first person's questions. (why amazon? Why physics --> software
: ?)
: The second person asked me 2 easy questions: reverse string and find al
y*i
19 楼
那就给孩子带帽子。在不影响司机后视的前提下，后座遮阳确实是有难度的。
【在 x***m 的大作中提到】
: 我们只有infant car seat有棚子，换成convertible后貌似只有一个座椅，没有棚子了
【在 x***m 的大作中提到】
: 我们只有infant car seat有棚子，换成convertible后貌似只有一个座椅，没有棚子了
j*l
20 楼
这个使用了pair的思想就是把原本也可以用递归函数实现的count一棵子树的节点总数，放到逆中序遍历的递归函数里头，共享同一个递归过程。这个思想同样可以用来求depth, 进而可以用来求二叉树的直径。好处都是无须再单独定义一个递归的count节点的函数和一个递归的求最大depth的函数，也无须把额外的节点个数信息或者depth信息保存到节点数据里。
need
enough
.
【在 c*********n 的大作中提到】
: could u share your result for below?
: But finally, I came back to the classical tree-algorithm. I
: almost got the correct answer -- with a little piece missing.
: basically the solution is very similar to in-order traversal: you just need
: to count how many nodes you have traversed so far. As soon as you got enough
: , you return that node. If you finished without reaching the number, you
: return null. You need to define a pair structure as the return
: value, where int represents the number o
need
enough
.
【在 c*********n 的大作中提到】
: could u share your result for below?
: But finally, I came back to the classical tree-algorithm. I
: almost got the correct answer -- with a little piece missing.
: basically the solution is very similar to in-order traversal: you just need
: to count how many nodes you have traversed so far. As soon as you got enough
: , you return that node. If you finished without reaching the number, you
: return null. You need to define a pair
: value, where int represents the number o
x*m
21 楼
我家娃特别痛恨帽子，哈哈，在商店给她试戴了N顶，每次都被一把抓下来，再戴就打挺
【在 y****i 的大作中提到】
: 那就给孩子带帽子。在不影响司机后视的前提下，后座遮阳确实是有难度的。
【在 y****i 的大作中提到】
: 那就给孩子带帽子。在不影响司机后视的前提下，后座遮阳确实是有难度的。
z*y
22 楼
cong!! well done
c*b
23 楼
能不能给后窗贴膜？我家的后窗有颜色，好像就不晒。
【在 x***m 的大作中提到】
: 宝宝朝后面坐着，有时太阳很大，直接晒到脸上，大家都单买遮太阳的吗？我没看到有
: 卖的啊
【在 x***m 的大作中提到】
: 宝宝朝后面坐着，有时太阳很大，直接晒到脸上，大家都单买遮太阳的吗？我没看到有
: 卖的啊
x*r
24 楼
Cong! 发包子吧！
a*e
25 楼
最近有啥convertible car seat的deal没？质量一般就行，这个作为backup用：）！
【在 x***m 的大作中提到】
: 宝宝朝后面坐着，有时太阳很大，直接晒到脸上，大家都单买遮太阳的吗？我没看到有
: 卖的啊
【在 x***m 的大作中提到】
: 宝宝朝后面坐着，有时太阳很大，直接晒到脸上，大家都单买遮太阳的吗？我没看到有
: 卖的啊
f*5
26 楼
I see
u want to use recursion to get the result.
while my solution is just use stack to do in-order traverse.
void traverse(Node* root)
{
Node *current = root;
Stack stack;
while (current!=NULL || !stack.empty)
{
if (current != NULL)
{
if (!stack.empty() && current->left == stack.top())
{counter++;
if (counter==n) { printf; return;}
current=null;
}
else { stack.push(current->left)
【在 c*********n 的大作中提到】
: could u share your result for below?
: But finally, I came back to the classical tree-algorithm. I
: almost got the correct answer -- with a little piece missing.
: basically the solution is very similar to in-order traversal: you just need
: to count how many nodes you have traversed so far. As soon as you got enough
: , you return that node. If you finished without reaching the number, you
: return null. You need to define a pair structure as the return
: value, where int represents the number o
u want to use recursion to get the result.
while my solution is just use stack to do in-order traverse.
void traverse(Node* root)
{
Node *current = root;
Stack stack;
while (current!=NULL || !stack.empty)
{
if (current != NULL)
{
if (!stack.empty() && current->left == stack.top())
{counter++;
if (counter==n) { printf; return;}
current=null;
}
else { stack.push(current->left)
【在 c*********n 的大作中提到】
: could u share your result for below?
: But finally, I came back to the classical tree-algorithm. I
: almost got the correct answer -- with a little piece missing.
: basically the solution is very similar to in-order traversal: you just need
: to count how many nodes you have traversed so far. As soon as you got enough
: , you return that node. If you finished without reaching the number, you
: return null. You need to define a pair
: value, where int represents the number o
j*l
27 楼
这个方法虽然直接，但却是O(n)的
This problem could be solved in O(log n) time if the BST contained the
number of nodes under a root
如果不允许把子树元素个数信息放到node中, 可以把算子树元素个数的递归算法，跟查
找kth node的递归算法合并在一起, 共享一个递归过程。复杂度是O(log n)
【在 f*********5 的大作中提到】
: I see
: u want to use recursion to get the result.
: while my solution is just use stack to do in-order traverse.
: void traverse(Node* root)
: {
: Node *current = root;
: Stack stack;
: while (current!=NULL || !stack.empty)
: {
: if (current != NULL)
This problem could be solved in O(log n) time if the BST contained the
number of nodes under a root
如果不允许把子树元素个数信息放到node中, 可以把算子树元素个数的递归算法，跟查
找kth node的递归算法合并在一起, 共享一个递归过程。复杂度是O(log n)
【在 f*********5 的大作中提到】
: I see
: u want to use recursion to get the result.
: while my solution is just use stack to do in-order traverse.
: void traverse(Node* root)
: {
: Node *current = root;
: Stack stack;
: while (current!=NULL || !stack.empty)
: {
: if (current != NULL)
s*l
28 楼
算子树个数的算法 也是o(n)啊~
怎么就logn的？
【在 j**l 的大作中提到】
: 这个方法虽然直接，但却是O(n)的
: This problem could be solved in O(log n) time if the BST contained the
: number of nodes under a root
: 如果不允许把子树元素个数信息放到node中, 可以把算子树元素个数的递归算法，跟查
: 找kth node的递归算法合并在一起, 共享一个递归过程。复杂度是O(log n)
怎么就logn的？
【在 j**l 的大作中提到】
: 这个方法虽然直接，但却是O(n)的
: This problem could be solved in O(log n) time if the BST contained the
: number of nodes under a root
: 如果不允许把子树元素个数信息放到node中, 可以把算子树元素个数的递归算法，跟查
: 找kth node的递归算法合并在一起, 共享一个递归过程。复杂度是O(log n)
j*l
29 楼
你说的对，我已经改了。
只有直接从节点里取出子树元素个数信息的情形下，才可以是O(log n)的。相当于空间
换时间并且作了预处理。
【在 s********l 的大作中提到】
: 算子树个数的算法 也是o(n)啊~
: 怎么就logn的？
只有直接从节点里取出子树元素个数信息的情形下，才可以是O(log n)的。相当于空间
换时间并且作了预处理。
【在 s********l 的大作中提到】
: 算子树个数的算法 也是o(n)啊~
: 怎么就logn的？
n*r
30 楼
没看懂，啥叫“放到逆中序遍历的递归函数里头，共享同一个递归过程”？
recursive去count node number，难到不是这样写：
int count(node *root)
{
if (!root)
return 0;
return (1+ count(root->left) + count(root->right);
}
本来就无须单独定义一个递归的count节点的函数。
数，放到逆中序遍历的递归函数里头，共享同一个递归过程。这个思想同样可以用来求
depth, 进而可以用来求二叉树的直径。好处都是无须再单独定义一个递归的count节点
的函数和一个递归的求最大d
【在 j**l 的大作中提到】
: 这个使用了pair的思想就是把原本也可以用递归函数实现的count一棵子树的节点总数，放到逆中序遍历的递归函数里头，共享同一个递归过程。这个思想同样可以用来求depth, 进而可以用来求二叉树的直径。好处都是无须再单独定义一个递归的count节点的函数和一个递归的求最大depth的函数，也无须把额外的节点个数信息或者depth信息保存到节点数据里。
:
: need
: enough
: .
recursive去count node number，难到不是这样写：
int count(node *root)
{
if (!root)
return 0;
return (1+ count(root->left) + count(root->right);
}
本来就无须单独定义一个递归的count节点的函数。
数，放到逆中序遍历的递归函数里头，共享同一个递归过程。这个思想同样可以用来求
depth, 进而可以用来求二叉树的直径。好处都是无须再单独定义一个递归的count节点
的函数和一个递归的求最大d
【在 j**l 的大作中提到】
: 这个使用了pair的思想就是把原本也可以用递归函数实现的count一棵子树的节点总数，放到逆中序遍历的递归函数里头，共享同一个递归过程。这个思想同样可以用来求depth, 进而可以用来求二叉树的直径。好处都是无须再单独定义一个递归的count节点的函数和一个递归的求最大depth的函数，也无须把额外的节点个数信息或者depth信息保存到节点数据里。
:
: need
: enough
: .
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