資料結構 Data Structures. Hsiao-Lung Chan, Ph.D. Dept Electrical Engineering Chang Gung University, Taiwan

Transcription

1 資料結構 Data Structures Hsiao-Lung Chan, Ph.D. Dept Electrical Engineering Chang Gung University, Taiwan

2 Topics Basic topics 1. Basic concepts 2. Arrays and Structures 3. Stacks and Queues 4. Linked lists 5. Trees 6. Graphs 7. Sorting 8. Hashing Advanced topics 1. Priority queues 2. Efficient binary search tree 3. Multiway search trees 4. Digital search trees Basic concepts 2

3 Textbook Ellis Horowitz, Sartaj Sahni, and Susan Anderson-Freed. Fundamentals of Data Structures in C, Silicon Press Basic concepts 3

4 How to create programs Requirement Analysis bottom-up vs. bottom-up Design data objects & operations Refinement & coding Verification proving, testing & debugging Basic concepts 4

5 Algorithm specification ( 演算法規格 ) An algorithm is a finite set of instructions that accomplishes a particular task. Criteria input output definiteness Each instruction is clear and unambiguous finiteness Algorithm terminates after a finite number of steps effectiveness Instruction is basic enough to be carried out Basic concepts 5

6 Data types Basic data type char, int, float, double Array int list[10]; Structure structure student { char lastname[20]; char studentid[10]; int age; Abstract data type (ADT) Basic concepts 6

7 Abstract data type An ADT is a data type organized in such a way that the specification of the objects and the operations is separated from the representation of the objects and the implementation of the operations. Basic concepts 7

8 Abstract data type: NaturalNumber structure NaturalNumber is objects: an ordered subrange of the integers starting at zero and ending at the maximum integer (INT_MAX) on the computer functions: for all x, y NaturalNumber; TRUE, FALSE Boolean and where +, -, <, and == are the usual integer operations. NaturalNumber Zero ( ) ::= 0 Boolean Is_Zero(x) ::= if (x) return FALSE else return TRUE NaturalNumber Add(x, y) ::= if ((x+y) <= INT_MAX) return x+y Boolean Equal(x,y) else return INT_MAX ::= if (x== y) return TRUE else return FALSE NaturalNumber Successor(x) ::= if (x == INT_MAX) return x else return x+1 NaturalNumber No Subtract(x,y) ::= if (x<y) return 0 else return x-y end Natural_Number Basic concepts 8

9 NaturalNumber implementation #define INT_MAX typedef structure { unsigned short val; NaturalNumber; NaturalNumber Add(NaturalNumber x, NaturalNumber y) { int sum; NaturalNumber z; void main(void) { sum=x.val+y.val; if sum>int_max NaturalNumber a,b,c; z.val=int_max; a.val=50; else z.val=(short) sum; b.val=60; return z; c=add(a,b); Basic concepts 9

10 Performance analysis (machine independent) Space complexity Storage requirement Time complexity Computing time Basic concepts 10

11 Space complexity: S(P)=C+S P (I) Fixed space requirements (C) Independent of the characteristics of the inputs and outputs instruction space space for simple variables, fixed-size structured variable, constants Variable space requirements (S P (I)) depend on the instance characteristic I number, size, values of inputs and outputs associated with I recursive stack space, formal parameters, local variables, return address Basic concepts 11

12 Example: A simple arithmetic function float abc(float a, float b, float c) { return a + b + b * c + (a + b - c) / (a + b) ; S abc (I) = 0 Basic concepts 12

13 Example: Iterative function for summing a list of numbers float sum(float list[ ], int n) { float tempsum = 0; int i; for (i = 0; i<n; i++) tempsum += list [i]; return tempsum; S sum (I) = 0 Recall: pass the address of the first element of the array & pass by value Basic concepts 13

14 Example: Recursive function for summing a list of numbers float rsum(float list[ ], int n) { if (n) return rsum(list, n-1) + list[n-1]; return 0; S sum (I) = 12n Space needed for one recursive call Type Name Number of bytes parameter: float list [ ] 4 parameter: integer n 4 return address:(used internally) 4 TOTAL per recursive call 12 Basic concepts 14

15 Time complexity: T(P)=C+T P (I) Compile time (C) independent of instance characteristics Run (execution) time T P It is not easy because it requires a detailed knowledge of compiler s attributes. For example, TP(n) = c a ADD(n) + c s SUB(n) + c l DA(n) + c st STA(n) Program step A syntactically or semantically meaningful program segment whose execution time is independent of the instance characteristics Basic concepts 15

16 Example: Iterative summing of a list of numbers float sum(float list[ ], int n) { float tempsum = 0; count++; /* for assignment */ int i; for (i = 0; i < n; i++) { count++; /*for the for loop */ tempsum += list[i]; count++; /* for assignment */ count++; /* last execution of for */ count++; return tempsum; /* for return */ 2n + 3 steps Basic concepts 16

17 Tabular method Statement steps/execution s/e Frequency Total steps float sum(float list[ ], int n) { float tempsum = 0; int i; for(i=0; i <n; i++) 1 n+1 n+1 tempsum += list[i]; 1 n n return tempsum; Total 2n+3 Basic concepts 17

18 Example: Recursive summing of a list of numbers float rsum(float list[ ], int n) { count++; /*for if conditional */ if (n) { count++; /* for rsum invocation */ return rsum(list, n-1) + list[n-1]; count++; /* for return */ return list[0]; 2n + 2 steps Basic concepts 18

19 Tabular method Statement s/e Frequency Total steps float rsum(float list[ ], int n) { if (n) 1 n+1 n+1 return rsum(list, n-1)+list[n-1]; 1 n n return list[0]; Total 2n+2 Basic concepts 19

20 Asymptotic notation (O) Definition (Big Oh ) f(n) = O(g(n)) if and only if there exist positive constants c and n 0 such that f(n) cg(n) for all n, n n 0. Examples 3n+2=O(n) /* 3n+2 4n for n 2 */ 3n+3=O(n) /* 3n+3 4n for n 3 */ 100n+6=O(n) /* 100n+6 101n for n 10 */ 10n 2 +4n+2=O(n 2 ) /* 10n 2 +4n+2 11n 2 for n 5 */ 6*2 n +n 2 =O(2 n ) /* 6*2 n +n 2 7*2 n for n 4 */ Basic concepts 20

21 Function values log n n nlog n n 2 n 3 2 n , ,536 4,294,967,296 Basic concepts 21

22 Plot of function values 60 n 2 2 n nlog n f n n log n Basic concepts 22

23 Complexity of c 1 n 2 +c 2 n and c 3 n For sufficiently large of value c 3 n is faster than c 1 n 2 +c 2 n For small values of n, either could be faster c 1 =1, c 2 =2, c 3 =100 --> c 1 n 2 +c 2 n c 3 n for n 98 c 1 =1, c 2 =2, c 3 = > c 1 n 2 +c 2 n c 3 n for n 998 Basic concepts 23

24 Performance measurement (machine dependent) Clocking Start timing Stop timing Type returned Result in seconds clock_t Method 1 start = clock(); stop = clock(); duration = ((double) (stop-start))/ CLOCKS_PER_SEC; start = time(null); stop = time(null); time_t Method 2 duration = (double) difftime(stop,start); Basic concepts 24

25 Timing program for selection sort #include <stdio.h> #include <time.h> #include selectionsort.h #define MAX_SIZE 1001 void main(void) { int i, n, step = 10, a[max_size]; double duration; clock_t start ; /* times for n = 0, 10,, 100, 200,, 1000 * / printf( n time\n ) ; for (n = 0; n <= 1000; n += step) { /*Initialize with worse case data * / for ( i = 0 ; i < n ; i++) a[i] = n i ; start = clock( ) ; sort(a, n) ; duration = ((double) (clock() start)) / CLOCKS_PER_SEC; printf( %6d %f\n, n, duration); if (n == 100) step = 100; Basic concepts 25

26 More accurate timing program for selection sort void main(void) { int i, n, step = 10, a[max_size]; double duration; printf( n repetitions time\n ); for (n = 0; n <= 1000; n += step) { long repetitions = 0 ; clock_t start = clock() ; do { repetitions++ ; for (i=0 ; i<n ; i++) a[i] = n i; sort(a, n) ; while(clock() start < 1000 ) /* repeat until enough time is elapsed * / duration = ((double) (clock() start)) / CLOCKS_PER_SEC; duration /= repetitions; printf( %6d %9d %f\n, n, repetitions,duration); if (n == 100) step = 100; Basic concepts 26