text: # 第 3 章 查找和排序
  
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  ### Text of 第 3 章 查找和排序
  
  • 3 3.1 3.2 3.3 3.4
  
  • 3.1 DS ASL ,
  
  • 3.2 --. 1) 1 2) ;,n,
  
  • 3.2 --. () : :1) mid =(left + right)/2 2) , , 3)
  
  • {3,5,11,17,21,23,28,30,32,50},30 1: {3,5,11,17,21,23,28,30,32,50} mid1= 1+10/2 = 5 2 { 2328303250 } mid2 = 6+10/2 = 8 : ,left
  
  • 3.2 --. 1. 1) nm(mn) 2) 3) 4)
  
  • 2. 1) 2) 3.
  
  • {22,12,13,9,8,33,42,44,38,24,48,60,58,75,47}1) : 2) [22,44,74],3) 60
  
  6044~743;3
  
  44
  
  22
  
  74
  
  2212139833424438244860587447List[1]List[2]List[3]
  
  • 3.3
  
  • 1) 2) 3) 4)
  
  • : {10183491325}:1)25: 25101825,32)35 101825 10394181325
  
  • 3.4 . K(),: Addr=HK : :,:
  
  • .
  
  • .1. :, K1K2 HK1= HK22. 3.
  
  • Hi =(H(key)+di)MOD m i=1,2,,k(k m-1) m:,di: 1) di=1,2,,m-1 2) di=12 ,-12,22,-22, ,+k2,-k2 (k m/2)
  
  • ,
  
  H(41)=H(1)=1
  
  H(67)=3
  
  H(530=H(13)=5
  
  H(22)=H(46)=H(30)=6 1
  
  2
  
  3
  
  4
  
  5
  
  6
  
  7
  
  41 1 NULL 46 13 NULL 30 NULL22NULL53NULLNULLNULL67
  
  • . {22,41,53,46,30,13,1,67},9: H(key) = key MOD 8 , Hi=(H(key)+di) MOD m1) H(22=6
  
  2) H(41=1 0 1 2 3 4 5 6 7 82222 0 1 2 3 4 5 6 7 841 1 1
  
  • 3) H(53= 5
  
  4) H(46= 6 Hi=(6+1MOD 8 = 7
  
  2241 0 1 2 3 4 5 6 7 8
  
  53224153460 1 2 3 4 5 6 7 8
  
  1 1 1 1 1 1 2
  
  • 5) H(30= 6 Hi=(6+1)MOD 8=7 Hi=0
  
  6) H(13= 522410 1 2 3 4 5 6 7 8
  
  53224153460 1 2 3 4 5 6 7 8
  
  3 1 1 1 2 3 1 6 1 1 246303013
  
  • 7) H(1= 1
  
  8) H(67= 322410 1 2 3 4 5 6 7 8
  
  5322415346 46303013 3 1 6 3 1 1 2113167 3 1 6 3 2 1 1 20 1 2 3 4 5 6 7 8
  
  • . HST[M]Rx
  
  (k)1. Di=HkHST[i]HST[i]=k
  
  2 Dk=R(Dk-1)HST[Dk]HST[Dk}=k
  
  • {22,41,53,46,30,13,1,67},H(key)=key MOD 8,,:
  
  ASL= (3+1+6+3+2+1+1+2)/8=19/8 key = 67 ,; key = 21 8,0 1 2 3 4 5 6 7 8
  
  3 1 6 3 2 1 1 2
  
  224153463013167
  
  • 13: 235~711 13
  
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explanation: The provided text appears to be a section from a textbook or lecture notes on data structures and algorithms, specifically focusing on searching and sorting techniques. The content is written in Chinese and covers topics such as binary search, hash functions, and collision resolution methods. The text includes several examples and step-by-step explanations of algorithms.
  
  The document does not contain a single, clear question or a question-and-answer pair. Instead, it provides detailed explanations and examples of various algorithms and techniques.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 第 3 章 查找和排序
  
  • View
  62
  
  0
  
  Embed Size (px)
  
  DESCRIPTION
  
  ### Text of 第 3 章 查找和排序
  
  • 3 3.1 3.2 3.3 3.4
  
  • 3.1 DS ASL ,
  
  • 3.2 --. 1) 1 2) ;,n,
  
  • 3.2 --. () : :1) mid =(left + right)/2 2) , , 3)
  
  • {3,5,11,17,21,23,28,30,32,50},30 1: {3,5,11,17,21,23,28,30,32,50} mid1= 1+10/2 = 5 2 { 2328303250 } mid2 = 6+10/2 = 8 : ,left
  
  • 3.2 --. 1. 1) nm(mn) 2) 3) 4)
  
  • 2. 1) 2) 3.
  
  • {22,12,13,9,8,33,42,44,38,24,48,60,58,75,47}1) : 2) [22,44,74],3) 60
  
  6044~743;3
  
  44
  
  22
  
  74
  
  2212139833424438244860587447List[1]List[2]List[3]
  
  • 3.3
  
  • 1) 2) 3) 4)
  
  • : {10183491325}:1)25: 25101825,32)35 101825 10394181325
  
  • 3.4 . K(),: Addr=HK : :,:
  
  • .
  
  • .1. :, K1K2 HK1= HK22. 3.
  
  • Hi =(H(key)+di)MOD m i=1,2,,k(k m-1) m:,di: 1) di=1,2,,m-1 2) di=12 ,-12,22,-22, ,+k2,-k2 (k m/2)
  
  • ,
  
  H(41)=H(1)=1
  
  H(67)=3
  
  H(530=H(13)=5
  
  H(22)=H(46)=H(30)=6 1
  
  2
  
  3
  
  4
  
  5
  
  6
  
  7
  
  41 1 NULL 46 13 NULL 30 NULL22NULL53NULLNULLNULL67
  
  • . {22,41,53,46,30,13,1,67},9: H(key) = key MOD 8 , Hi=(H(key)+di) MOD m1) H(22=6
  
  2) H(41=1 0 1 2 3 4 5 6 7 82222 0 1 2 3 4 5 6 7 841 1 1
  
  • 3) H(53= 5
  
  4) H(46= 6 Hi=(6+1MOD 8 = 7
  
  2241 0 1 2 3 4 5 6 7 8
  
  53224153460 1 2 3 4 5 6 7 8
  
  1 1 1 1 1 1 2
  
  • 5) H(30= 6 Hi=(6+1)MOD 8=7 Hi=0
  
  6) H(13= 522410 1 2 3 4 5 6 7 8
  
  53224153460 1 2 3 4 5 6 7 8
  
  3 1 1 1 2 3 1 6 1 1 246303013
  
  • 7) H(1= 1
  
  8) H(67= 322410 1 2 3 4 5 6 7 8
  
  5322415346 46303013 3 1 6 3 1 1 2113167 3 1 6 3 2 1 1 20 1 2 3 4 5 6 7 8
  
  • . HST[M]Rx
  
  (k)1. Di=HkHST[i]HST[i]=k
  
  2 Dk=R(Dk-1)HST[Dk]HST[Dk}=k
  
  • {22,41,53,46,30,13,1,67},H(key)=key MOD 8,,:
  
  ASL= (3+1+6+3+2+1+1+2)/8=19/8 key = 67 ,; key = 21 8,0 1 2 3 4 5 6 7 8
  
  3 1 6 3 2 1 1 2
  
  224153463013167
  
  • 13: 235~711 13
  
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text appears to be a section from a textbook or lecture notes on data structures and algorithms, specifically focusing on searching and sorting techniques. The content is written in Chinese and covers topics such as binary search, hash functions, and collision resolution methods. The text includes several examples and step-by-step explanations of algorithms.
  
  The document does not contain a single, clear question or a question-and-answer pair. Instead, it provides detailed explanations and examples of various algorithms and techniques.
  
  Label: no-question