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T*********s
发帖数: 20444 | 2 再送给你一个吧
Toyota Avalon XLE
1) Truecar Price：under 28K
2) IIHS测试：全优5个(G) TSP
3) 0~60mph: 6.0s （ZEROto60times数据， 2013model）
4）至少20年历史，质量非常好，口碑极佳。
5）Plus: 安全质量豪华保值形象，样样第一。 |
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a*****3
发帖数: 10373 | 3 这是旧款的c63 coupe, 新款的sedan刚开始卖，加速是3.9/4.0s，大概7，8万多。新款
的coupe刚公布，明年卖，加速3.9s, 起价大概8万。 |
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s********u
发帖数: 748 | 4 Truecar是税前，另外不建议2.0S，115马力实在是带不动 |
|
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t******2
发帖数: 2265 | 6 楼上的童靴们，我说的是GLC 63 S Coupe，2018才上市，不是GLC 63,或GLC 63 S，那
不是一个级别的。
这个比较的实质是，Levante S，5.0s的肉启动，装B，还是GLC 63 S Coupe，AMG-
503HP，3.7s的启动灭法拉利F430半条街，就别说他干儿子了。
所以是装B,还是注重实质?这是个问题。
至于楼上那个C的童鞋，暴露了装B的人生。GLC首先是G，其次是L,然后才是C。C，E，S
跟宝马1，3，5，7一样是区别车身尺寸大小的标识，要知道GLC 63 S Coupe比GLE 63 S
Coupe还要快10%，尴尬了吧，伐哈哈哈~
呕，要知道GLE的前身是ML,臭名昭著的名声，还是在阿拉巴马产的，谁还敢买？当然
coupe是个新系列，分散在各个分支里了。GLC在德国产的，得多脑残的装B才会去卖GLE
哈。 |
|
T**********e
发帖数: 387 | 7 不懂别装懂。 只有coupe才有glc63s。suv只有glc63，没有glc63s。
: 楼上的童靴们，我说的是GLC 63 S Coupe，2018才上市，不是GLC 63,或GLC 63
S，那
: 不是一个级别的。
: 这个比较的实质是，Levante S，5.0s的肉启动，装B，还是GLC 63 S Coupe，
AMG-
: 503HP，3.7s的启动灭法拉利F430半条街，就别说他干儿子了。
: 所以是装B,还是注重实质?这是个问题。
: 至于楼上那个C的童鞋，暴露了装B的人生。GLC首先是G，其次是L,然后才是C。C
，E，S
: 跟宝马1，3，5，7一样是区别车身尺寸大小的标识，要知道GLC 63 S Coupe比
GLE 63 S
: Coupe还要快10%，尴尬了吧，伐哈哈哈~
: 呕，要知道GLE的前身是ML,臭名昭著的名声，还是在阿拉巴马产的，谁还敢买？
当然
: coupe是个新系列，分散在各个分支里了。GLC在德国产的，得多脑残的装B才会
去卖GLE
|
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l*****6
发帖数: 7881 | 8 我自己的途锐就是买的cpo，帮同事买过两台cpo。奔驰砍了1K，Audi砍了2k，我自己的
途锐砍了2k。我经常关注cpo豪车市场。一般豪车cpo都可以有1，2k的降价空间。
Motortrend测得audi a6 2.0t 0-60是6.0s, Avalon是6.2s, 四缸的怎么带不动a6了？
现在所有豪车都有四缸turbo的引擎选择。
24k
500 |
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l**********6
发帖数: 330 | 9 我想卖的物品:
1 个 Macbook Pro, 2012 Model, 13.3"/i5/500G 5400转/4G内存/0S X Lion
型号：MD313LL/A
单张面值:
$1305 (retail$1199+tax$106)
可接受价格(必须明码标价！):
$1050
物品新旧:
NEW (没有盒子，原因是买了本子后手快把盒子扔了，但是本子外壳上苹果的塑料封套
和安全封条没有打开，本子完全全新，没有打开或者是用过，所有说明书和电源等等都
是完好无损，没有打开或用过。）
苹果 warranty 到 2013年1月。
http://newyork.craigslist.org/que/sys/2790317876.html
邮寄方式要求:
N/A (NEW YORK/ New Jersey local only)
付款方式说明:
cash
我的联系方式:
d********[email protected]
state and zip:
New York, 11373 |
|
c*********o
发帖数: 399 | 10 二手交易风险自负！请自行验证是否合法和一手卡！:
我想卖的物品:
Iphone 3gs 16gb+ipad wifi 16gb
单张面值:
iphone 3gs black (0S 3.1.3 or 3.3.1) used working condition great.无附件，$
310+YL
ipad 16gb new sealed in box $515+YL
可接受价格(必须明码标价！):
物品新旧要求:
邮寄方式要求:
买卖双方谁承担邮寄损失(Required if not code only):
付款方式说明:
non cc paypal
其他补充说明:
广告的有效期:
物品来源(Required for All Cards!):
我的联系方式:
Warranty期限:
能否证明是合法的一手卡?(Required for All Cards!):
state and zip: |
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x**i
发帖数: 88 | 11 The effects list says:
No:12
SMALL ROOM 3
approx. 1.0s reverb decay |
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x******e
发帖数: 13 | 12 bitmap should give you a O(n) solution if all numbers are integer.
1. find min and max of all the input integers. O(n)
2. define a binary array in the range of [min, max] (or binary map)
3. when reading each pair of integers [a, b], set the bits between a and b
to 1 if they are 0s. - O(n)
4. finally, scan the bitmap and output the continuous region. O(n).
I didn't implement it, but it should work for integer inputs. If any number
is not integer, the method doesn't work.
Any comments? |
|
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c***z
发帖数: 6348 | 14 What I finally arrive at before timed out was this:
if there is a 1 on a line or row, change that line or row to all 1;
for each blocks of 0s, min{height, width} is the dimension of the submatrix.
two passes on the matrix, O(MN). |
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b****r
发帖数: 1272 | 15 sum method will suffer overflow issues and may no tbe what interviewers are
looking for
for 1 missing number, one can use bitwise (2^n-1) searching - count # of 0s
and 1s at each digit
for >1 missing numbers, binary search is one of the approaches |
|
b********e
发帖数: 693 | 16 Given a NxN matrix with 0s and 1s. Now whenever you encounter a 0 make the
corresponding row and column elements 0.
Flip 1 to 0 and 0 remains as they are.
for example
1 0 1 1 0
0 1 1 1 0
1 1 1 1 1
1 0 1 1 1
1 1 1 1 1
results in
0 0 0 0 0
0 0 0 0 0
0 0 1 1 0
0 0 0 0 0
0 0 1 1 0
my solution is define a function remove1(x1,y1,x2,y2)
(x1,y1) is the matrix start point, and (x2,y2) is the end point
if(i,j) == 1, we need remove all 1 from i row and j column
so we divide into four small matrix
resrucive |
|
UD
发帖数: 182 | 17
example
this is your example:
1100101101010111111111111
I am getting this result:
>./array_find_subseq
0 0 1 0 1 1 0 1 0 1 0 1
>Exit code: 0
it is one of the right answers.
the answer in your previous post, seems not right, there should be 6 0s.
>>the longest subarray is 1001011010. in this case, j should moves right!
and there is no guarantee, you can find j in O(1) time. |
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h**k
发帖数: 3368 | 18 How about this case? 10 1s and 11 0s
110001110000011100011
space
and |
|
i**********e
发帖数: 1145 | 19 这题蛮有意思的，我刚写完。
其实 idea 挺容易明白，我说一次给你听就明白了，但是没图解释起来比较费劲。这题
最复杂的地方其实就是选择怎么把数据结构结合起来。
一开始我以为要用 dp，其实 greedy 就可以了。
总复杂度是 O(N lg M)，N 为 str 的长度，M 为 pattern 的长度。
主要原因有个 lg M 是因为 STL map 里的 find() 函数复杂度为 O(lg M).
我用的是 map + queue + hashtable （有点吓人呵呵，可能我想太复杂了）。
我暂时还没想到怎么提升到 O(N)，应该是利用一个更好的数据结构吧。如果有高人知
道怎么提升到 O(N)，请指点一下吧~
这是我做的 test cases:
第一行是 string 和 pattern，
第二行是函数 return 的 start and end position，然后是 shortest substring。
cabeca cae
3 5 eca
cfabeca cae
4 6 eca
cabefgecdaecf cae
9 11 aec
cabwefgewcw... 阅读全帖 |
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g***s
发帖数: 3811 | 20 since we dont care the stablility of 0, so, we can use the 0 to adjust.
if the (number of 0) / (number of array) is a constant level, it can be
solved by O(n).
assum number of 0 is k
1. Partition (or compact,or something like inplace delete) to non-0 and 0 O
(n)
2. check if number of position < number of negative. pick smaller one to do
3-5. assume we choose positive number.
3. scan from end, non-0 number;
4. swap k position number with 0s
5. compact 0, go to 4, util all positive numbers are mo... 阅读全帖 |
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g***s
发帖数: 3811 | 21 or do you have better solution? My solution is just using the k 0s as a
extra space. I mentioned in another post, if k<
be used. |
|
t*******i
发帖数: 4960 | 22 我上机试了一下，
std::cout << f(100) << endl;
std::cout << f(10) << endl;
std::cout << f(22010) << endl;
637534208
1342177280
1604976640
for example v = 10. its binary form
becomes 010100000....(28 0s) |
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b***e
发帖数: 15201 | 23 前面贴出来的解法还有个Bug,数组前几行都是0的时候，返回值差1：
这个是改正的版本。
public static int countComps(int[][] a)
{
int m=a.length;//row length
int n=a[0].length; //col length
int label=1;//label init value is 1 but will start with 2,increase
by 1 when find 1 possible new component
HashSet equal_labels = new HashSet();//to save equivalent label
pairs,no duplicates
//return result would be
int west=0;
int north=0;
... 阅读全帖 |
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z**********8
发帖数: 229 | 24 career cup5.1
题目：
You are given two 32-bit numbers, N and M, and two bit positions, i and j.
Write a
method to set all bits between i and j in N equal to M (e.g., M becomes a
substring of
N located at i and starting at j).
EXAMPLE:
Input: N = 10000000000, M = 10101, i = 2, j = 6
Output: N = 10001010100
答案：
This code operates by clearing all bits in N between position i and j, and
then ORing to put
M in there.
1 public static int updateBits(int n, int m, int i, int j) {
2 int max = ~0; /* All 1’s... 阅读全帖 |
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l******0
发帖数: 313 | 26 trading position, 总的来说题目挺难的，主要是时间很短，虽然在interviewer的提
醒下，最终找到正确答案，但是刚开始的答案是错的。。。不知道还能不能进入下一轮
了~
1, 10 digit number, 1st digit representing the number of 0s, 2nd digit
representing the number of 1s, ..., 10th digit representing the number of 9s
. What are the possible numbers?
2, 2 dies. 6 sided and 8 sided. play game. you guess the sum of the two dies
. you get the same number of dollars if you guess correct. what number will
you choose to maximize your profit? |
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p*****2
发帖数: 21240 | 27 The way Microsoft’s review system works, a whole bunch of Microsoft
employees just got their annual performance reviews. The September 15
paystub will reflect their new “numbers”. If they got a merit increase
the paystub will show it. If they got a bonus, it will show up there too.
This post is for those Microsoft employees who are not happy with their
review…
That September 15 date is what motivates managers to finally get busy
delivering the bad news to employees who didn’t fare so well in s... 阅读全帖 |
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d******i
发帖数: 76 | 28 If overflow is not to be considered.
Then what you should do is get the multiplication of all elements.
But the tricky part is that.
1. 0 exists
2. how many 0s exist.
How to deal with the above two situations is the key to solve this problem.
If overflow is an issue.
Then you could use DP, very simple DP.
Two times scan. one from left ->right, one from right -> left.
left ->right iteration record the multiplication of elements to the left of
current element. ex. MUL_Left[i] = A[i] * MUL_Left[ i ... 阅读全帖 |
|
z********i
发帖数: 568 | 29 231: 2>1, 改1为2。中间的3为回文。返回。
999： 9=9，中间的为回文，但没法增加，改两端的9为10&01，中间设为位数减1全0--
〉1001。
9999：--〉10001
989: 9=9，中间的为回文，但可增加，--->999
假设没有leading 0s
用两个函数：
1。isP（int x): return true如果x是回文，否则false.
2. isMaxP(int x): return true如果x=99...9。
find_next_palindrome_int(ax..zb)
if(isMaxP(x..z) or x..z is empty)
if(a==9)
if(b==9)
x..z=0..0
return 10x..z01
else
return ax...z9
else
if(b>=a)
a=a+1;
b=a;
else
b=a;
else
... 阅读全帖 |
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c**i
发帖数: 306 | 30 可以的，但是001010这样的是不合法的，不能有leading 0s |
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r*********n
发帖数: 4553 | 31 idea:
first use rand5() to generate a binary random function with p = 1/7 as
follows
express 1/7 in base 5
e.g. 0.abc......
(Note if x can be expressed in base 5 with finite digits, after the last
digit, we assume an infinite number of 0s are padded)
check first digit a:
if rand5() < a return 1
else if rand5() > a return 0
else proceed to the next digit
similarly generate binary random functions with p=1/6, 1/5, ..... 1/2
denote these binary random functions as rand(), N = 2,3,...,7
if rand<7... 阅读全帖 |
|
c******3
发帖数: 60 | 32 谢谢大家的bless。今天总体还是算顺利的没有见到新的算法题（庆幸？)，就不贴了。
比如tree non-recursive iteration (inorder), interval merge and insert,
binary matrix coloring (0->1, 1->0 except 1s surrounded by 0s)
不过 open ended 的system design的问题一大堆，不好过关。
比如：
how would you design youtube?
do you know about GFS? what could you do to improve it? and why?
what would you improve gmail? why?
How would you design counters for youtube videos? |
|
c******3
发帖数: 60 | 33 谢谢大家的bless。今天总体还是算顺利的没有见到新的算法题（庆幸？)，就不贴了。
比如tree non-recursive iteration (inorder), interval merge and insert,
binary matrix coloring (0->1, 1->0 except 1s surrounded by 0s)
不过 open ended 的system design的问题一大堆，不好过关。
比如：
how would you design youtube?
do you know about GFS? what could you do to improve it? and why?
what would you improve gmail? why?
How would you design counters for youtube videos? |
|
c***f
发帖数: 40 | 34 binary matrix coloring (0->1, 1->0 except 1s surrounded by 0s)
请问这道题有什么好的解法吗， 在网上也没搜到这道题，
哪位大牛能解答下，或给个链接呢 |
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c***w
发帖数: 134 | 35 这题怎么做啊？？
You are given a n*n matrix of bits (1s and 0s) where 1 represents land and 0
represents water. Adjacent 1s can be considered as joined together to form
sort of island in water. Count the number of islands |
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l*****c
发帖数: 52 | 36 前前后后面了四个月 电面面了两次 第一次onsite结束以后 Hiring Committee要求加
面一次两轮的onsite 周五HR电话打来说offer还是不能approve 决定move-on 接着面别
的家了
攒人品 求refer
Phone 1st
1. Hash table (collision, probability)
2. Generate fuzzy words (Not in dictionary, but look like the given
strings)
implement build() and nextWord();
e.g. ["APE", "APPLE", "ACE"] -> "ACE" (randomly)
Phone 2nd
1. Lowest common ancestor
2. Can't remember
Onsite 1st
1 White Female Mira
1.1 Find intersection from two lists
1.2 How many 0... 阅读全帖 |
|
j*t
发帖数: 184 | 37 >1.2 How many 0s tailing in N!
这题是数1...N中末尾数是5或0的个数吗？
>2.1 10 buttons passcode with 4 digitals. Generate a sequence to
> brute force it. Upper bound and lower bound of length, code to
> generate an optimal one.
>
>If we start from 0000 and use minimum digits if the number doesn't >appear
previously, we get 0000100020003000400050006000700080009 and >it will form a
cycle and stop generating new numbers.
backtracking? 如何证明soultion接近10k+3? |
|
s****p
发帖数: 124 | 38 You meant to convert 0 to -1, right? Otherwise, it won't work, because for i
< j, if there are same numbers of 1s and 0s between them, C[i] and C[j] won
't map to the same bucket, because C[j] = C[i] + (j-i)/2. |
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b********6
发帖数: 97 | 41 背景：本科生物，统计master + 9个月工作经验
结果: offer: amazon, facebook, linkedin, google
Withdraw了ebay的onsite，别的好多电面都fail或者没有消息
电面：
Amazon两个：面得太早，具体想不起来了，code题不多。问怎么从某种格式的log file
里抓出想要的信息，简单的regular expression 和perl scripts, 问一些如果server
有问题怎么trouble shooting的开放问题。
Linkedin 两个：
1 binary tree level order traversal, leetcode原题
2 pow(x,2) leetcode原题
3 判断一个string表示的数字是否valid，类似leetcode Valid Number原题，一些具体
要求要和面试官讨论后确定
4 permutation I and II leetcode原题
Facebook一个：
1 reverse linkedlist (这个我无话可说)
2 decide whether tw... 阅读全帖 |
|
g******y
发帖数: 143 | 42 Consider N coins aligned in a row. Each coin is showing either heads or
tails. The adjacency of these coins is the number of adjacent pairs of coins
with the same side facing up.
It must return the maximum possible adjacency that can be obtained by
reversing exactly one coin (that is, one of the coins must be reversed).
Consecutive elements of array A represent consecutive coins in the row.
Array A contains only 0s and/or 1s: 0 represents a coin with heads facing up
; 1 represents a coin with ta... 阅读全帖 |
|
a******d
发帖数: 82 | 43 面试中的coding 题目都答出来了, 虽然被指出几个小bug, 但是当时也立即就改过来了.
design 的题目也照自己复习的内容答了, 不是很有经验,但是对new grads 也就这个水
平了
最后还是收到被拒的邮件, 自己都不知道是怎么一回事.
Facebook onsite 问到的几个题目
1. moving all 0s to the beginning of the array
直接答出 constant space 的解法
2. strstr()
KMP
3. wordbreak
告诉面试官有DP 解法, 告知我写递归. 我也写出个递归解法, 并且用hashmap 保存出
现过的子串来优化. |
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c**********t
发帖数: 8 | 44 纠结它只让改三行code, 毫无头绪，求解
Consider N coins aligned in a row. Each coin is showing either heads or
tails. The adjacency of these coins is the number of adjacent pairs of coins
with the same side of face.
You are giving an implementation of a function:
int solution(int A[], int N);
that, given a non-empty zero-idexed array A consisting of N integers
representing the coins, returns the maximum possible adjacency that can be
obtained by reversing exactly one coin (that is, one of... 阅读全帖 |
|
m*****g
发帖数: 71 | 45 JUnit test 代码
public class SolutionTest {
@Test
public void test() {
int[] a;
int[] b;
boolean expect;
Solution s = new Solution();
//both empty
a = new int[] {};
b = new int[] {};
expect = true;
testSolution(s, a, b, expect);
//empty and 1 element
a = new int[] {};
b = new int[] {1};
expect = true;
testSolution(s, a, b, expect);
//empty and 2 element all 0s
a =... 阅读全帖 |
|
a********5
发帖数: 1631 | 46 组里被阿三垄断。我的直属MANAGER A，第二等级MANAGER B, DIRECTOR C. 全是阿三。
A是从今年一月开始变成MANAGER TRACK。 B很提拔A，B本来是我的直属MANAGER，基本
没管过我，除了扔活的时候，1:1 能有1/3按时发生就不错，提拔A做MANAGER之后更是
直接就把WEEKLY的1:1变成MONTHLY的。C是去年空降的，交流很少。第一次交流就是下
述的这件事。
A喜欢PUSH 乱七八糟的活给我。B有一段时间也有活给我，他们之间不协调导致我的工
作进度很混乱，有一段时间我同时有四五个杂七杂八的TASK在推进。这其中有一个小
FEATURE，C亲自PUSH，因为他想要给CEO做DEMO，所以有一天过来直接问我多久能做完
（第一次交流）。A坐我附近，我还什么都没说，A就说他一天就可以做完。C就说那好
，明天结束之前我要看到。然后就走了。
所以我就开始做，结果当天傍晚时B突然找我，要我做另一个修改。我说这是一个不小
的修改，可能需要半天到一天时间，我可以做但是我需要等到A,C布置的项目做完后再
开始。B说：他们的项目你先放一放，那个东西不可能给C... 阅读全帖 |
|
x**********g
发帖数: 44 | 47 x,y 是int吗？如果是，唯一能想到的用matrix（all 0s），扫一个个rectangles 变1
，最后数1的个数~~当然实在太暴力了 |
|
b*****n
发帖数: 618 | 48 前段时间骑驴找马终于告一段落，感觉本版的技术贴和面经贴帮助非常之大，也非常感
谢共享资源的各路大牛。希望提供一些信息和个人感受给还在找工的童鞋，有帮助最好
，但是毕竟本人资历尚浅，如果有不对的地方也请轻喷。
背景：
ms毕业不到两年
主要申请公司：
offer：facebook，google，uber，palantir，sumo logic，walmartlab，yahoo，
amazon，apple
reject：dropbox
主要几个包裹：
U： 145k base + 25k股 RSU
F： 150k base + 40k signon + 10%bonus + 260k美元 RSU
W： 165k base + 50k signon + 20%bonus + 35k美元 RSU每年（
这个略复杂，相当于每年35k美元RSU的refresh，但是每次refresh分四年给）
再上各个公司的面经和感受：
Yahoo：
最早面的公司，面的是Flurry Team，Yah... 阅读全帖 |
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f*******r
发帖数: 976 | 49 恭喜，都是好包袱！
关键字: 面经
发信站: BBS 未名空间站 (Sat Jun 13 17:27:31 2015, 美东)
前段时间骑驴找马终于告一段落，感觉本版的技术贴和面经贴帮助非常之大，也非常感
谢共享资源的各路大牛。希望提供一些信息和个人感受给还在找工的童鞋，有帮助最好
，但是毕竟本人资历尚浅，如果有不对的地方也请轻喷。
背景：
ms毕业不到两年
主要申请公司：
offer：facebook，google，uber，palantir，sumo logic，walmartlab，yahoo，
amazon，apple
reject：dropbox
主要几个包裹：
U： 145k base + 25k股 RSU
F： 150k base + 40k signon + 10%bonus + 260k美元 RSU
W： 165k base + 50k signon + 20%bonus + 35k美元 RSU每年（
这个略复杂，相当于每年35k美元RSU的refres... 阅读全帖 |
|
g*****9
发帖数: 4125 | 50 Yes, they know how this game is played.
Lots of chinese people work hard to make money, but put them in the bank or
invest them, so people like 好莱坞明星 can borrow them and really enjoy life.
I am not sure this is sad or not, but life is too short to just die with
bunch of 0s to your name.
Enjoy while you are still young to have some fun. |
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