# 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

Recommended

Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents