d*******4
发帖数: 43 | 1 第一轮：老美
0. a couple questions regarding resume
1. polymorphism, abstraction
2. reverse words in a given string (coding)
3. what is hash table, collision resolution methods
4. 设计电梯
read code in phone for 2
第二轮：老美
interviewer had problem with his laptop, so gave questions out of memory
0. Java final, finally, finalize
1. what is hash table
2. reverse an integer (in binary form)
I gave O(n) solution.
3. 三连击问题
4. design a card game
asked to send codes for 2 and 4 before 9AM the next morning.
I had a couple small mistakes, but were pointed out by interviewers.
I directly asked both interviewers feedback at the end. They both gave
positive feedback. And the recruiter told me to go to next around one day
after interview.
They do like to ask about Hash table. Make sure you know the details. |
B******5
发帖数: 4676 | |
r********r
发帖数: 2912 | 3 What does that mean for "reverse an integer in binary form"?
【在 d*******4 的大作中提到】
: 第一轮：老美
: 0. a couple questions regarding resume
: 1. polymorphism, abstraction
: 2. reverse words in a given string (coding)
: 3. what is hash table, collision resolution methods
: 4. 设计电梯
: read code in phone for 2
: 第二轮：老美
: interviewer had problem with his laptop, so gave questions out of memory
: 0. Java final, finally, finalize
|
d*******4
发帖数: 43 | 4 if you have n = 10 (1010 in binary), then reverse it, you get 5 (0101 in
binary).
【在 r********r 的大作中提到】
: What does that mean for "reverse an integer in binary form"?
|
i**9
发帖数: 351 | 5 我以为是n=10 binary 0x0000000a reverse以后是 0x50000000
【在 d*******4 的大作中提到】
: if you have n = 10 (1010 in binary), then reverse it, you get 5 (0101 in
: binary).
|
s*******3
发帖数: 134 | 6 bless!
顺便问一下你第二轮电面之后第二天就收到recruiter的onsite invitation拉？谢谢 |
s*******3
发帖数: 134 | 7 bless!
顺便问一下你第二轮电面之后第二天就收到recruiter的onsite invitation拉？谢谢 |
d*******4
发帖数: 43 | 8 应该是第三天，interviewer第二天收到程序才交的报告吧。
【在 s*******3 的大作中提到】
: bless!
: 顺便问一下你第二轮电面之后第二天就收到recruiter的onsite invitation拉？谢谢
|
q*******g
发帖数: 110 | |
w**********u
发帖数: 172 | |
|
|
m****i
发帖数: 650 | |
h**********d
发帖数: 4313 | 12 不错，我二面后仅一个月才收到onsite，nonlocal就是慢啊 |
f*****w
发帖数: 2602 | |
n********p
发帖数: 708 | 14 bless~~~~~~~~~~~~~~~~~~~ |
g*******s
发帖数: 2963 | 15 2.2 楼主说的On是指 bit位数么？
可不可以这么做:
x = 5;
reverse(x){
y = 0;
for(int i = 0; i
x = 1 & y
y<<1
x>>1
return y;
} |
a******7
发帖数: 106 | 16 about 3. 三连击问题
could any one make the problem clear, the three hits should belong to one
user or not?
I searched around, this question has been repeated many times. But I didn't
find clear problem definition.
Thanks! |
d*******4
发帖数: 43 | 17 given many log files of webpage visits with user id and time stamp, find out
what is the most popular three-page sequence of all. Assume log files
totally are too big for the main memory to hold.
The general idea is to hash on user id, sort on time stamp, and then hash
three-page sequence to store how many times they appear.
t
【在 a******7 的大作中提到】
: about 3. 三连击问题
: could any one make the problem clear, the three hits should belong to one
: user or not?
: I searched around, this question has been repeated many times. But I didn't
: find clear problem definition.
: Thanks!
|
d*******4
发帖数: 43 | 18 Yes.
The best solution is O(log n) using bit mask. Someone has posted the solution here a while ago.
The page below gives an example when the integer is 32 bits. But it can be generalized to n bits:
http://aggregate.org/MAGIC/#Bit%20Reversal
【在 g*******s 的大作中提到】
: 2.2 楼主说的On是指 bit位数么？
: 可不可以这么做:
: x = 5;
: reverse(x){
: y = 0;
: for(int i = 0; i: x = 1 & y
: y<<1
: x>>1
: return y;
|
s******f
发帖数: 3984 | |
j********x
发帖数: 2330 | |
h****b
发帖数: 25 | 21 big bless! 随便问下楼主你的第二次店面题目挺多的， 不是都要写code吧？ 还是大
部分只是说解法了。 要不然感觉45分钟肯定不够吧！！ |
d*******4
发帖数: 43 | 22 只说算法，代码店面后写完发给面试官。
【在 h****b 的大作中提到】
: big bless! 随便问下楼主你的第二次店面题目挺多的， 不是都要写code吧？ 还是大
: 部分只是说解法了。 要不然感觉45分钟肯定不够吧！！
|
