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
  
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  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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