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池大侠 发表于 2015-3-17 02:01:10 | 显示全部楼层 |阅读模式

2013(1-3月) 码农类 硕士 全职@Google - 内推 - 技术电面 |Other

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just finished .. all three can be solve with hashing..... from: 1point3acres.com/bbs
Bless!!
鏉ユ簮涓浜.涓夊垎鍦拌鍧.
first question:
1.find the latest events
2.find pairs in hands  need to define data structure by myself
3.find straight in hands

all problem can be done in linear time.
below are questions and my solution.

bless!!!!!!!!!!! anyway....
  1. Please use this Google doc to code during your interview. To free your hands for coding, we recommend that you use a headset or a phone with speaker option.

  2. Best,
  3. Google Staffing
  4. .鐣欏璁哄潧-涓浜-涓夊垎鍦

  5. A session is a series of Events. Each session is identified by a number, the sessionId. Events have this structure:

  6. class Event {.1point3acres缃
  7.   public int sessionId;  // identities the session this Event belongs to. 1point3acres.com/bbs
  8.   public int timestamp;  // time the Event occurred. 1point 3acres 璁哄潧
  9.   [...]
  10. }

  11. We have an unordered list of Events from multiple sessions intermixed. Given this list (List<Event>), construct a new list containing only the latest Event for each session.
  12. #input=[event1,event2,.....]
  13. #linear space and linear time
  14. #time1 12:00PM
  15. #hashtable
  16. def solve(events):
  17.         dict={}# key: sessionID, value would be eventname sorted by timestamp
  18. for elem in events:       
  19.                 if elem.sessionId not in dict:
  20.                         dict[elem.sessionId]=elem
  21.                 elif elem.timestamp>dict[elem.sessionId].timestamp:. Waral 鍗氬鏈夋洿澶氭枃绔,
  22.                         dict[elem.sessionId]=elem
  23. res=[]
  24. for elem in dict:. 1point3acres.com/bbs
  25.         res.append(dict[elem])
  26. return res


  27. . 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴

  28. 52-card deck
  29. 4 suits: spades, hearts, diamonds, clubs
  30. 13 values: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. from: 1point3acres.com/bbs
  31.        

  32. hand: 5 cards. 涓浜-涓夊垎-鍦帮紝鐙鍙戝竷
  33. Write a function to detect whether a poker hand contains a “pair” (exactly 2 cards with the same value).
  34. ## given 5 cards

  35. class card:
  36.         def __init__(self,value,suit):
  37.                 self.value=value
  38.                 self.suit=suit. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
  39. #class hand:. From 1point 3acres bbs
  40. #        def __init__(self,n):
  41. #                self.n=n
  42. #                self.hand=[]
    . Waral 鍗氬鏈夋洿澶氭枃绔,
  43. #        def getcard(self,card):
  44. #                self.hand.append(card)
  45.         . From 1point 3acres bbs
  46. #list of cards=>hand
  47. #nlog(n) time sort by its value  give me the result 3, 8 eight?
  48. #hash. From 1point 3acres bbs
  49. # linear time and space
  50. def check(hand):
  51.         dict={}
  52.         for card in hand:
  53.                 if card.value not in dict:
    . 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
  54.                         dict[card.value]=1
  55.                 else:
  56.                         dict[card.value]+=1
  57.         ## check whether  I have a pair
  58.         for elem in dict:
  59.                 if dict[elem]==2:
  60.                         return True
  61.         return False. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴

  62. # determine whether the hand contains a “straight”: values are consecutive
  63. # Ace can be high or low. from: 1point3acres.com/bbs
  64. # A, 2, 3, 4, 5 is a straight
  65. # 4, 5, 6, 7, 8 is a straight
  66. # 10, J, Q, K, A is also a straight       
  67. # replace A with both 1 and 14
  68. # sort nlog(n)
  69. # 7,8,9,10,A -> 7,8,9,10,1,14
  70. # 10,11,12,13,A-> 10,11,12,13,14,1
  71. # go through the sorted array of cards to check whether they are consecutive
  72. # this take nlogn time
  73. # hash to detect consecutive cards



  74. . 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴

  75. def check(hand):
  76.         ## deal with A first -> replace A with 1 and 14. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
  77.         ## ‘J’:11, ‘K’:13####. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
  78. . from: 1point3acres.com/bbs
  79. for cardelem in hand:
  80.         ## first step deal with A
  81. if cardelem.value=’A’:.1point3acres缃
  82.         newcard=card(14,’spade’)
  83.         cardelem.value=1
  84. hand.append(newcard)
  85. dict={}
  86. . from: 1point3acres.com/bbs
  87. for cardelem in dict:
  88.         if cardelem not in dict:
  89.                 dict[cardelem.value]=True       
  90.                      count=1
  91.                 maxlen=1.1point3acres缃
  92.                 for key in dict:
  93.                         count=1
  94.                         pre=key-1
  95.                         while pre in dict and dict[pre]:
  96.                                 count+=1
  97.                                 dict[pre]=False
  98.                                 pre-=1
  99.         ## 6,7<-8 ->9 10. 1point3acres.com/bbs
  100.                         post=key+1
  101.                         while post in dict and dict[post]:
  102.                                 count+=1
  103.                                 dict[post]=False
  104.                                 post+=1
  105.                         maxlen=max(maxlen,count)
  106.         if maxlen==5:. Waral 鍗氬鏈夋洿澶氭枃绔,
  107.                 return True
  108.         else:        . 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
  109.                 return False
  110.                
  111. #linear time and space
  112. #
  113. . visit 1point3acres.com for more.
  114.        
  115. Ace of Spades
  116. 3 of diamonds
  117. 8 of clubs
  118. 8 of hearts
  119. J of diamonds
复制代码

补充内容 (2015-3-16 17:43):
some  problem in the code fomat cause jst copied from google doc.

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jeager 发表于 2015-3-17 02:45:54 | 显示全部楼层
关注一亩三分地公众号:
Warald_一亩三分地
第三题 可以 这么办
一个boolean A, A是否出现
一个max,一个min
A不计入max,min,length
一圈下来
if(A){
    if((min == 2 || max == 13) && max-min +1 == 4) return true;
    else return false;
}
else{
  if(max-min +1 == 5) return true;
  else return false;
}
. visit 1point3acres.com for more.
这样的话,time O(n), space(1);

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fishyuze 发表于 2015-3-18 09:21:01 | 显示全部楼层
关注一亩三分地微博:
Warald
for the straight: why not just use a boolean array?
and what about J Q K A 2?
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houqingniao 发表于 2015-3-17 02:36:26 | 显示全部楼层
bless~
lz 没问题的
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Reborn2beCoder 发表于 2015-3-18 07:25:15 | 显示全部楼层
跪求lz解释这段代码

              for key in dict:
                        count=1. 1point 3acres 璁哄潧
                        pre=key-1. Waral 鍗氬鏈夋洿澶氭枃绔,
                        while pre in dict and dict[pre]:
                                count+=1
                                dict[pre]=False
                                pre-=1
        ## 6,7<-8 ->9 10.鐣欏璁哄潧-涓浜-涓夊垎鍦
                        post=key+1. 1point 3acres 璁哄潧
                        while post in dict and dict[post]:
                                count+=1
                                dict[post]=False
                                post+=1.1point3acres缃
                        maxlen=max(maxlen,count)
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 楼主| 池大侠 发表于 2015-3-18 08:47:09 | 显示全部楼层
Reborn2beCoder 发表于 2015-3-17 22:25
跪求lz解释这段代码

              for key in dict:

longest consecutive sequence...
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Reborn2beCoder 发表于 2015-3-18 08:56:46 | 显示全部楼层
池大侠 发表于 2015-3-18 08:47
longest consecutive sequence...

可是pre 跟 post 都只加一次就被false了呀。。还会继续运行吗><
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 楼主| 池大侠 发表于 2015-3-18 09:22:43 | 显示全部楼层
Reborn2beCoder 发表于 2015-3-17 23:56
可是pre 跟 post 都只加一次就被false了呀。。还会继续运行吗>

yes.. just need to check once . if they are consecutive it doesnt matter.
that's y its linear
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Reborn2beCoder 发表于 2015-3-18 09:37:50 | 显示全部楼层
池大侠 发表于 2015-3-18 09:22.鏈枃鍘熷垱鑷1point3acres璁哄潧
yes.. just need to check once . if they are consecutive it doesnt matter.
that's y its linear

看明白啦~~~ 不过其实如果只是连着的5个数的话只要判断没有重复而且dic[4] - dic[0] == 4就好了吧?有A的话就看 dic[4] - dic[0] ==4 || dic[5]-dic[1]==4 ?
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