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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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x
just finished .. all three can be solve with hashing....
Bless!!

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. . more info on 1point3acres.com
  3. Best,
  4. Google Staffing


  5. A session is a series of Events. Each session is identified by a number, the sessionId. Events have this structure:. 1point 3acres 璁哄潧

  6. class Event {
  7.   public int sessionId;  // identities the session this Event belongs to
  8.   public int timestamp;  // time the Event occurred
  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.鐣欏璁哄潧-涓浜-涓夊垎鍦
  21.                 elif elem.timestamp>dict[elem.sessionId].timestamp:
  22.                         dict[elem.sessionId]=elem
  23. res=[]
  24. for elem in dict:
  25.         res.append(dict[elem])
  26. return res


  27. .鐣欏璁哄潧-涓浜-涓夊垎鍦

  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
  31.        

  32. hand: 5 cards
  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. visit 1point3acres.com for more.
  38.                 self.suit=suit.鐣欏璁哄潧-涓浜-涓夊垎鍦
  39. #class hand:
  40. #        def __init__(self,n):. from: 1point3acres.com/bbs
  41. #                self.n=n
  42. #                self.hand=[]
  43. #        def getcard(self,card):
  44. #                self.hand.append(card)
  45.        
  46. #list of cards=>hand. 鍥磋鎴戜滑@1point 3 acres
  47. #nlog(n) time sort by its value  give me the result 3, 8 eight?
  48. #hash
  49. # linear time and space
  50. def check(hand):
  51.         dict={}. 1point3acres.com/bbs
  52.         for card in hand:
  53.                 if card.value not in dict:
  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

  62. # determine whether the hand contains a “straight”: values are consecutive. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
  63. # Ace can be high or low
  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. from: 1point3acres.com/bbs
  71. # go through the sorted array of cards to check whether they are consecutive.鐣欏璁哄潧-涓浜-涓夊垎鍦
  72. # this take nlogn time
  73. # hash to detect consecutive cards
  74. .鏈枃鍘熷垱鑷1point3acres璁哄潧



  75. . 1point 3acres 璁哄潧
  76. def check(hand):
  77.         ## deal with A first -> replace A with 1 and 14
  78.         ## ‘J’:11, ‘K’:13####

  79. for cardelem in hand:
  80.         ## first step deal with A
  81. if cardelem.value=’A’:
  82.         newcard=card(14,’spade’). more info on 1point3acres.com
  83.         cardelem.value=1
  84. hand.append(newcard)
  85. dict={}
  86. . From 1point 3acres bbs
  87. for cardelem in dict: 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
  88.         if cardelem not in dict:
  89.                 dict[cardelem.value]=True       
  90.                      count=1
  91.                 maxlen=1
  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 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
  100.                         post=key+1. Waral 鍗氬鏈夋洿澶氭枃绔,
  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:
  107.                 return True. From 1point 3acres bbs
  108.         else:       
  109.                 return False
  110.                
  111. #linear time and space 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
  112. #

  113.        
  114. Ace of Spades
  115. 3 of diamonds
  116. 8 of clubs
  117. 8 of hearts. more info on 1point3acres.com
  118. 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
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;.鐣欏璁哄潧-涓浜-涓夊垎鍦
    else return false;
}
else{. From 1point 3acres bbs
  if(max-min +1 == 5) return true;
  else return false;
}. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴

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
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
.鐣欏璁哄潧-涓浜-涓夊垎鍦
              for key in dict:. Waral 鍗氬鏈夋洿澶氭枃绔,
                        count=1
                        pre=key-1
                        while pre in dict and dict[pre]:
                                count+=1
                                dict[pre]=False
                                pre-=1
        ## 6,7<-8 ->9 10
                        post=key+1. 1point 3acres 璁哄潧.1point3acres缃
                        while post in dict and dict[post]:-google 1point3acres
                                count+=1
                                dict[post]=False
                                post+=1
                        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. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
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
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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