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Data Structures
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The logical or mathematical model of a particular organization of data is called a data structure
DATA STRUCTURES
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A primitive data type holds a single piece of data–e.g. in Java: int, long, char, boolean etc.–Legal operations on integers: + - * / ...
A data structure structures data!–Usually more than one piece of data–Should provide legal operations on the data–The data might be joined together (e.g. in an array): a collection
DATA STRUCTURES
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Static vs. Dynamic StructuresStatic vs. Dynamic Structures
A static data structure has a fixed size
This meaning is different than those associated with the static modifier
Arrays are static; once you define the number of elements it can hold, it doesn’t change
A dynamic data structure grows and shrinks as required by the information it contains
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An Abstract Data Type (ADT) is a data type together with the operations, whose properties are specified independently of any particular implementation.
Abstract Data Type
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Abstract Data Type
In computing, we view data from three perspectives: Application level
View of the data within a particular problem Logical level
An abstract view of the data values (the domain) and the set of operations to manipulate them
Implementation level A specific representation of the structure to hold the data
items and the coding of the operations in a programming language
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Problem Solving: Main Steps
1. Problem definition2. Algorithm design / Algorithm specification3. Algorithm analysis4. Implementation5. Testing6. [Maintenance]
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Problem Definition
What is the task to be accomplished?Calculate the average of the grades for a given student
What are the time / space / speed / performance requirements?
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. Algorithm Design / Specifications
Algorithm: Finite set of instructions that, if followed, accomplishes a particular task.Describe: in natural language / pseudo-code / diagrams / etc. Criteria to follow:
Input: Zero or more quantities (externally produced)Output: One or more quantities Definiteness: Clarity, precision of each instructionFiniteness: The algorithm has to stop after a finite (may be very large) number of stepsEffectiveness: Each instruction has to be basic enough and feasible
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Implementation, Testing, Maintenances
ImplementationDecide on the programming language to use
C, C++, Lisp, Java, Perl, Prolog, assembly, etc. , etc.Write clean, well documented code
Test, test, test
Integrate feedback from users, fix bugs, ensure compatibility across different versions Maintenance
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Algorithm Analysis
Space complexityHow much space is required
Time complexityHow much time does it take to run the algorithm
Often, we deal with estimates!
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Space Complexity
Space complexity = The amount of memory required by an algorithm to run to completion
[Core dumps = the most often encountered cause is “memory leaks” – the amount of memory required larger than the memory available on a given system]
Some algorithms may be more efficient if data completely loaded into memory
Need to look also at system limitationsE.g. Classify 2GB of text in various categories [politics, tourism, sport, natural disasters, etc.] – can I afford to load the entire collection?
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Space Complexity (cont’d)
1. Fixed part: The size required to store certain data/variables, that is independent of the size of the problem:- e.g. name of the data collection- same size for classifying 2GB or 1MB of texts
2. Variable part: Space needed by variables, whose size is dependent on the size of the problem:- e.g. actual text - load 2GB of text VS. load 1MB of text
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Space Complexity (cont’d)
S(P) = c + S(instance characteristics)c = constant
Example:float sum (float* a, int n) {
float s = 0; for(int i = 0; i<n; i++) { s+ = a[i]; } return s;}Space? one word for n, one for a [passed by reference!], one for i constant space!
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Time Complexity
Often more important than space complexityspace available (for computer programs!) tends to be larger and largertime is still a problem for all of us
3-4GHz processors on the market researchers estimate that the computation of various transformations for 1 single DNA chain for one single protein on 1 TerraHZ computer would take about 1 year to run to completion
Algorithms running time is an important issue
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Running Time
Problem: prefix averagesGiven an array XCompute the array A such that A[i] is the average of elements X[0] … X[i], for i=0..n-1
Sol 1At each step i, compute the element X[i] by traversing the array A and determining the sum of its elements, respectively the average
Sol 2 At each step i update a sum of the elements in the array ACompute the element X[i] as sum/I
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Running time
Input
1 ms
2 ms
3 ms
4 ms
5 ms
A B C D E F G
worst-case
best-case}average-case?
Suppose the program includes an if-then statement that may execute or not: variable running timeTypically algorithms are measured by their worst case
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Experimental Approach
Write a program that implements the algorithmRun the program with data sets of varying size.Determine the actual running time using a system call to measure time (e.g. system (date) );
Problems?
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Experimental Approach
It is necessary to implement and test the algorithm in order to determine its running time.
Experiments can be done only on a limited set of inputs, and may not be indicative of the running time for other inputs.
The same hardware and software should be used in order to compare two algorithms. – condition very hard to achieve!
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Use a Theoretical Approach
Based on high-level description of the algorithms, rather than language dependent implementations
Makes possible an evaluation of the algorithms that is independent of the hardware and software environments
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Algorithm Description
How to describe algorithms independent of a programming language Pseudo-Code = a description of an algorithm that is
more structured than usual prose but less formal than a programming language
(Or diagrams)Example: find the maximum element of an array.
Algorithm arrayMax(A, n):Input: An array A storing n integers.Output: The maximum element in A.currentMax A[0]for i 1 to n -1 do
if currentMax < A[i] then currentMax A[i]return currentMax
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Properties of Big-Oh
Expressions: use standard mathematical symbols use for assignment ( ? in C/C++)use = for the equality relationship (? in C/C++)
Method Declarations: -Algorithm name(param1, param2) Programming Constructs:
decision structures: if ... then ... [else ..]while-loops while ... do repeat-loops: repeat ... until ... for-loop: for ... do array indexing: A[i]
Methodscalls: object method(args)returns: return value
Use commentsInstructions have to be basic enough and feasible!
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Asymptotic analysis - terminology
Special classes of algorithms:logarithmic: O(log n)linear: O(n)quadratic: O(n2)polynomial: O(nk), k ≥ 1exponential: O(an), n > 1
Polynomial vs. exponential ?Logarithmic vs. polynomial ?
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Some Numbers
log n n n log n n2 n3 2n
0 1 0 1 1 21 2 2 4 8 42 4 8 16 64 163 8 24 64 512 2564 16 64 256 4096 655365 32 160 1024 32768 4294967296
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Relatives of Big-Oh
“Relatives” of the Big-Oh (f(n)): Big Omega – asymptotic lower bound (f(n)): Big Theta – asymptotic tight bound
Big-Omega – think of it as the inverse of O(n)g(n) is (f(n)) if f(n) is O(g(n))
Big-Theta – combine both Big-Oh and Big-Omegaf(n) is (g(n)) if f(n) is O(g(n)) and g(n) is (f(n))
Make the difference: 3n+3 is O(n) and is (n)3n+3 is O(n2) but is not (n2)
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More “relatives”
Little-oh – f(n) is o(g(n)) if for any c>0 there is n0 such that f(n) < c(g(n)) for n > n0.Little-omegaLittle-theta
2n+3 is o(n2) 2n + 3 is o(n) ?
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Example
Remember the algorithm for computing prefix averages compute an array A starting with an array X every element A[i] is the average of all elements X[j] with j < iRemember some pseudo-code … Solution 1Algorithm prefixAverages1(X):Input: An n-element array X of numbers.Output: An n -element array A of numbers such that A[i] is the average of elements X[0], ... , X[i].
Let A be an array of n numbers.for i 0 to n - 1 do
a 0for j 0 to i do
a a + X[j] A[i] a/(i+ 1)
return array A
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Example (cont’d)
Algorithm prefixAverages2(X):Input: An n-element array X of numbers.Output: An n -element array A of numbers such that A[i] is the average of elements X[0], ... , X[i].
Let A be an array of n numbers.s 0for i 0 to n do
s s + X[i] A[i] s/(i+ 1)
return array A
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Back to the original question
Which solution would you choose?O(n2) vs. O(n)
Some math …properties of logarithms:logb(xy) = logbx + logbylogb (x/y) = logbx - logbylogbxa = alogbxlogba= logxa/logxb–properties of exponentials:a(b+c) = aba c
abc = (ab)c
ab /ac = a(b-c)
b = a loga
b
bc = a c*loga
b
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Important Series
Sum of squares:
Sum of exponents:
Geometric series:Special case when A = 2
20 + 21 + 22 + … + 2N = 2N+1 - 1
N largefor 36
)12)(1( 3
1
2 NNNNiN
i
-1k and N largefor |1|
1
1
k
NikN
i
k
111
0
A
AANN
i
i
N
i
NNiNNS1
2/)1(21)(
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Analyzing recursive algorithms
function foo (param A, param B) {statement 1;statement 2;if (termination condition) {
return;foo(A’, B’);
}
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Solving recursive equations by repeated substitution
T(n) = T(n/2) + c substitute for T(n/2)= T(n/4) + c + c substitute for T(n/4)= T(n/8) + c + c + c= T(n/23) + 3c in more compact form= …= T(n/2k) + kc “inductive leap”
T(n) = T(n/2logn) + clogn “choose k = logn”= T(n/n) + clogn= T(1) + clogn = b + clogn = θ(logn)
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Solving recursive equations by telescoping
T(n) = T(n/2) + c initial equation T(n/2) = T(n/4) + c so this holds T(n/4) = T(n/8) + c and this … T(n/8) = T(n/16) + c and this …
… T(4) = T(2) + c eventually … T(2) = T(1) + c and this … T(n) = T(1) + clogn sum equations, canceling the terms appearing on both sides T(n) = θ(logn)
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RECURSION
Suppose P is a procedure containing either a CALL statement to itself or a CALL statement back to original procedure P .Then P is called a recursive procedure
Properties:
1. There must be certain criteria called basic criteria, for which the procedure does not call itself.
2. Each time the procedure does call itself (directly or indirectly), it must be closer to the base criteria.
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FACTORIAL WITHOUT RECURSION
FACTORIAL(FACT,N)
This procedure calculates N! and return the vale in the variable FACT .
1. If N ==0,then :Set FACT:=1, and Return.
2. Set FACT:=1[Initialize FACT for loop]
3. Repeat for K:=1 to N
Set FACT:=K*FACT
[END of loop]
4. Return.
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FACTORIAL WITH RECURSION
FACTORIAL(FACT,N)
This procedure calculates N! and return the vale in the variable FACT .
1. If N ==0,then :Set FACT:=1, and Return.
2. Call FACTORIAL(FACT,N-1).
3. Set FACT:=N*FACT.
4. Return.
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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FACTORIAL EXAMPLE USING RECURSION
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Stack
A stack is a list that has addition and deletion of items only from one end.
It is like a stack of plates:Plates can be added to the top of the stack.Plates can be removed from the top of the stack.
This is an example of “Last in, First out”, (LIFO).
Adding an item is called “pushing” onto the stack.
Deleting an item is called “popping” off from the stack.
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STACK OPERATION (PUSH)
PUSH(STACK,TOP,MAXSTK,ITEM)
This procedure pushes an ITEM onto a stack.
1.[Stack already filled]
If TOP== MAXSTK, then: Print:OVERFLOW, and Return.
2. Set TOP:=TOP+1.[ Increases TOP by 1]
3. Set STACK[TOP]:=ITEM. [Inserting ITEM in new TOP position]
4. Return.
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STACK OPERATION (POP)
POP(STACK,TOP,ITEM)
This procedure deletes the top element of STACK and assigns it to the variable ITEM .
1.[Stack has an item to be to removed]
If TOP== 0, then: Print:UNDERFLOW, and Return.
2. Set ITEM:=STACK[top].[ Assigns TOP element to ITEM ]
3. Set TOP:=TOP-1. [Decreases TOP by 1]
4. Return.
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STACK EXAMPLES
Implementing a stack :
MAXSTK=7.
Data
push 43
push 23
push 53
pop
pop
push 100
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STACK EXAMPLES
7
6
5
4
2
1
TOP=0
MAXSTK=7
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STACK EXAMPLES
7
6
5
4
2
1
TOP=1
MAXSTK=7
PUSH(43)
43
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STACK EXAMPLES
7
6
5
4
2
1
TOP=2
MAXSTK=7
PUSH(23)
43
23
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STACK EXAMPLES
7
6
5
4
2
1
TOP=3
MAXSTK=7
PUSH(53)
43
23
53
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STACK EXAMPLES
7
6
5
4
2
1
TOP=2
MAXSTK=7
POP( )
43
23
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STACK EXAMPLES
7
6
5
4
2
1
TOP=0
MAXSTK=7
POP( )
43
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STACK EXAMPLES
7
6
5
4
2
1
TOP=0
MAXSTK=7
PUSH(100)
43
100
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STACK EXAMPLES
1. INFIX to POSTFIX
2. POSTFIX expression solving
3. N-QUEEN’S problem
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INFIX to POSTFIX
Properties while transforming infix to postfix expression
1. besides operands and operators, arithmetic expression contains left and right parentheses
2. We assume that the operators in q consist only of
1. Exponent
2. Multiplication
3. Division
4. Addition
5. Subtraction
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INFIX to POSTFIX
3. We have three levels of precedence
precedence operators
high right parentheses
exponent
multiplication, division
low addition, subtraction
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INFIX to POSTFIX
POLISH(Q, P) Suppose Q is an arithmetic expression written in infix notation. This algorithm finds the equivalent postfix expression P.
1. Push “(“ into STACK and add “)” to the end of Q.
2. Scan Q from left to right and repeat Step 3 to 6 for each element of Q until the STACK is empty:
3. If an operand is encountered add it to P.
4. If a left parenthesis is encountered, push it onto STACK.
5. If an operator is encountered then:
(a) repeatedly pop from STACK and to P each operator (on the top of STACK) which has the same precedence as or higher precedence this operator
(b) Add operator to STACK
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INFIX to POSTFIX
6. If a right parenthesis is encountered then:
(a) Repeatedly pop from STACK and add to P each operator( on the top of STACK) until a left parenthesis is encountered
(b) Remove the left parenthesis .[Do not add the left parenthesis to P]
7. Exit
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INFIX to POSTFIX
INFIX Expression: 3 + 2 * 4POSTFIX Expression:
7
6
5
4
2
1
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INFIX to POSTFIX
INFIX Expression: + 2 * 4POSTFIX Expression: 3
7
6
5
4
2
1
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INFIX to POSTFIX
INFIX Expression: 2 * 4POSTFIX Expression: 3
7
6
5
4
2
1
+
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INFIX to POSTFIX
INFIX Expression: * 4POSTFIX Expression: 3 2
7
6
5
4
2
1
+
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INFIX to POSTFIX
INFIX Expression: 4POSTFIX Expression: 3 2
7
6
5
4
2
1
+
*
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INFIX to POSTFIX
INFIX Expression: POSTFIX Expression: 3 2 4
7
6
5
4
2
1
+
*
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INFIX to POSTFIX
INFIX Expression: POSTFIX Expression: 3 2 4 *
7
6
5
4
2
1
+
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INFIX to POSTFIX
INFIX Expression: POSTFIX Expression: 3 2 4 * +
7
6
5
4
2
1
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POSTFIX EXPRESSION SOLVING
This algorithm finds the VALUE of an arithmetic expression P written in postfix notation
1. Add a right parenthesis “)” at the end of P
2. Scan P from left to right and repeat step 3 and 4 for each element of P until the sentinel “)” is encountered.
3. If an operand is encountered, put it on STACK.
4. If an operator is uncounted then:
a. Remove the two elements of Stack, where A is the top element and B is the next top element
b. Evaluate B operator A
c. Pace the result of (b) back on STACK
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POSTFIX EXPRESSION SOLVING
5. Set VALUE equal to the top element on STACK
6. Exit
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POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 3 2 4 * +Push the numbers until operator comes 7
6
5
4
2
1
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POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 2 4 * +
7
6
5
4
2
1
3
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POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 4 * +
7
6
5
4
2
1
3
2
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77
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: * + Here we pop two time and perform multiplication on elements (4*2) and push the Result in to the stack
7
6
5
4
2
1
3
4
2
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POSTFIX EXPRESSION SOLVING
POSTFIX Expression: +
7
6
5
4
2
1
3
8
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79
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 3 2 4 * +
7
6
5
4
2
1
11
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The N-Queens Problem
Can the queens be placed on the board so that no two queens are attacking each other in chess board
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81
The N-Queens Problem
Two queens are not allowed in the same rowTwo queens are not allowed in the same row
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82
The N-Queens Problem
Two queens are not allowed in the same row, or in the same column...Two queens are not allowed in the same row, or in the same column...
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The N-Queens Problem
Two queens are not allowed in Two queens are not allowed in the same row, or in the same the same row, or in the same column, or along the same column, or along the same diagonal.diagonal.r along the same diagonal.
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84
The N-Queens Problem
The number of queens, and the size of the board can vary.
The number of queens, and the size of the board can vary.
N Queens
N columns
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The 3-Queens Problem
The program uses a The program uses a stack to keep track of stack to keep track of where each queen is where each queen is placed.placed.
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86
The 3-Queens Problem
Each time the program decides to place a queen on the board, the position of the new queen is stored in a record which is placed in the stack.
ROW 1, COL 1
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87
The 3-Queens Problem
We also have an integer variable to keep track of how many rows have been filled so far.
ROW 1, COL 1
1 filled
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The 3-Queens Problem
Each time we try to place a new queen in the next row, we start by placing the queen in the first column
ROW 1, COL 1
1 filled
ROW 2, COL 1
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The 3-Queens Problem
if there is a conflict with another queen, then we shift the new queen to the next column.
ROW 1, COL 1
1 filled
ROW 2, COL 2
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The 3-Queens Problem
ROW 1, COL 1
1 filled
ROW 2, COL 3
When there are no When there are no conflicts, we stop and conflicts, we stop and add one to the value of add one to the value of filled.filled.
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91
The 3-Queens Problem
When there are no conflicts, we stop and add one to the value of filled.
ROW 1, COL 1
2 filled
ROW 2, COL 3
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92
The 3-Queens Problem
Let's look at the third row. The first position we try has a conflict
ROW 1, COL 1
2 filled
ROW 2, COL 3
ROW 3, COL 1
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The 3-Queens Problem
so we shift to column 2. But another conflict arises
ROW 1, COL 1
2 filled
ROW 2, COL 3
ROW 3, COL 2
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94
The 3-Queens Problem
and we shift to the third column.Yet another conflict arises
ROW 1, COL 1
2 filled
ROW 2, COL 3
ROW 3, COL 3
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95
The 3-Queens Problem
and we shift to column 4. There's still a conflict in column 4, so we try to shift rightward again
ROW 1, COL 1
2 filled
ROW 2, COL 3
ROW 3, COL 4
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The 3-Queens Problem
but there's nowhere else to go.
ROW 1, COL 1
2 filled
ROW 2, COL 3
ROW 3, COL 4
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97
The 3-Queens Problem
When we run out of room in a row: pop the stack, reduce filled by 1 and continue
working on the previous row.
ROW 1, COL 1
1 filled
ROW 2, COL 3
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The 3-Queens Problem
Now we continue working on row 2, shifting the queen to the right.
ROW 1, COL 1
1 filled
ROW 2, COL 4
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99
The 3-Queens Problem
This position has no conflicts, so we can increase filled by 1, and move to row 3.
ROW 1, COL 1
2 filled
ROW 2, COL 4
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The 3-Queens Problem
In row 3, we start again at the first column.
ROW 1, COL 1
2 filled
ROW 2, COL 4
ROW 3, COL 1
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The 3-Queens Problem
In row 3, we start again at the first column.
ROW 1, COL 1
3 filled
ROW 2, COL 4
ROW 3, COL 2
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QUEUES
A queue is a data structure that maintains a “first-in first-out” (FIFO) ordering.
In contrast, a stack maintains a “last-in first-out” (LIFO) ordering.
A queue adds new elements at the end. An element can only be removed at the front.
This is an abstraction of the “first-come first-served” practice.
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QUEUE OPERATIONS
A queue has two operations:QINSERTQDELETE
An enqueue operation adds new elements at the end of the queue or its tail. This is similar to the stack operation push; only that push now is done at the end of the array instead of at the front (or top).
A dequeue operation removes an element from the front of the array or its head.
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QUEUE INSERTION
QINSERT(rear, item)
1. IF rear == MAX_QUEUE_SIZE_1 then Print queue_full Return;
2. INFO [++rear] = item;
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QUEUE DELETION
QDELETE(front, rear)
1. IF front == rear then Return queue_empty( ); Return queue [++ front];
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QUEUE
OFFSET
DATA
0 1 2 3
0
rear
0
OFFSET
DATA
0 1 2 3
A
front
1
rear
1
OFFSET
DATA
0 1 2 3
A B
front
1
rear
2
OFFSET
DATA
0 1 2 3
B
front
2
rear
2
front
Insert A
Insert B
delete
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CIRCULAR QUEUE
When a new item is inserted at the rear, the to rear moves upwards.
Similarly, when an item is deleted from the queue the front arrow moves downwards.
After a few insert and delete operations the rear might reach the end of the queue and no more items can be inserted although the items from the front of the queue have been deleted and there is space in the queue.
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CIRCULAR QUEUE
To solve this problem, queues implement wrapping around. Such queues are called Circular Queues.Both the front and the rear wrap around to the beginning of the array when they reached the MAX size of the queue.
It is also called as “Ring buffer”.
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CIRCULAR QUEUE INSERTION
QINSERT (QUEUE, N, FRONT, ITEM) This procedure insert an element ITEM into a queue.
1. [Queue already filled?]
IF ( FRONT==1 and REAR==N ) or FRONT ==REAR + 1,then: write: overflow, and Return
2.[Find new value of REAR] IF FRONT==NULL then [Queue initially empty.]
Set FRONT=1 and REAR=1
ELSE IF REAR ==N then Set REAR=1
ELSE set REAR=REAR+1
[End of if structure]
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CIRCULAR QUEUE INSERTION
3. Set QUEUE[REAR]=ITEM.[This inserts new element]
4. Return.
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CIRCULAR QUEUE DELETION
QDELETE(QUEE, N, FRONT, REAR, ITEM)
This procedure deletes an element from a queue and assigns it to the variable ITEM
1.[Queue already empty]
if FRONT=NULL then write UNDERFLOW, and Return
2. Set ITEM=QUEUE[FRONT]
3. [Find new value of FRONT]
If FRONT =REAR then [Queue has only one element to start]
Set FRONT=NULL and REAR=NULL
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CIRCULAR QUEUE DELETION
Else if FRONT ==N then
Set FRONT=1
Else
Set FRONT=FRONT +1
[End of if statement]
4.Return
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113
CIRCULAR QUEUE DELETION
EMPTY QUEUE[2] [3] [2] [3]
[1] [4] [1] [4]
[0] [5] [0] [5]
front = 0 front = 0 rear = 0 rear = 3
J2
J1
J3
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114
CIRCULAR QUEUE DELETION
[2] [3] [2] [3]
[1] [4][1] [4]
[0] [5] [0] [5]
front =0rear = 5
front =4rear =3
J2 J3
J1 J4
J5 J6 J5
J7
J8 J9
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PRIORITY QUEUE
A priority queue is a collection of elements such that each element has been assigned a priority and such that the order in which elements are deleted and processed comes from the following rules
1. An element of a higher priority is processed before any elements of lower priority
2. Two elements with the same priority are processed according to the order in which they were added to the queue
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ARRAY REPRESENTATION PRIORITY QUEUE
Use a separate queue for each level of priority
Each such queue will appear in its own circular array and must have its own pair of pointers FRONT and REAR
In fact, if each queue is allocated the same amount of space in two-dimensional array QUEUE can be used instead of linear array
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DELETION ON PRIORITY QUEUE
Deletion
This algorithm deletes and processes the first element in a priority queue maintained by a two-dimensional array QUEUE
1.[Find the first nonempty queue] Find the smallest k that FRONT!=NULL
2.Delete and process the front element in row K of QUEUE
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INSERTION ON PRIORITY QUEUE
Insertion
This algorithm adds an ITEM with priority number M to a priority queue maintained by a two-dimensional array QUEUE
1. Insert ITEM as the rear element in row M of QUEUE
2. Exit
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LINKED LIST
Linked list Linear collection of self-referential class objects, called nodesConnected by pointer linksAccessed via a pointer to the first node of the listSubsequent nodes are accessed via the link-pointer member of the current nodeLink pointer in the last node is set to null to mark the list’s end
Use a linked list instead of an array whenYou have an unpredictable number of data elementsYour list needs to be sorted quickly
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TYPES OF LINKED LIST
Types of linked lists:Singly linked list
Begins with a pointer to the first nodeTerminates with a null pointerOnly traversed in one direction
Circular, singly linkedPointer in the last node points back to the first node
Doubly linked listTwo “start pointers” – first element and last elementEach node has a forward pointer and a backward pointerAllows traversals both forwards and backwards
Circular, doubly linked listForward pointer of the last node points to the first node and backward pointer of the first node points to the last node
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linked list
Start NodeA placeholder node at the beginning of the list, used to simplify list processing. It doesn’t hold any data
Tail NodeA placeholder node at the end of the list, used to simplify list processing
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Singly linked list
Single Linked ListConsists of data elements and reference to the
next Node in the linked listFirst node is accessed by referenceLast node is set to NULL
A B CStart Tail
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SINGLE LINKED LIST INSERTION
INSFIRST(INFO, LINK, START, AVAIL, ITEM)
This algorithm inserts ITEM as the first node in the list
1. [OVERFLOW?] If AVAIL=NULL then :Write OVERFLOW and Exit
2. [Remove first node from AVIL list] Set NEW =AVAIL and AVAIL=LINK[AVAIL]
3. Set INFO[NEW]=ITEM [Copies new data into new node]
4. Set LINK[NEW]=START [new node now points to original first node]
5. Set START=NEW [Changes START so it points to the node]
6. Exit
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SINGLE LINKED LIST INSERTION
s X c d a
current
1
start
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SINGLE LINKED LIST INSERTION
s X c d a
current
1
start
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SINGLE LINKED LIST INSERTION
s X c d a
current
1
start
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SINGLE LINKED LIST INSERTION
s X c d a
current
1
start
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SINGLE LINKED LIST INSERTION
s X c d a
current
1
start
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SINGLE LINKED LIST INSERTION
INSLOC(INFO, LINK,START, AVAIL, LOC, ITEM) This algorithm inserts ITEM so that ITEM follows the node with location LOC or inserts ITEM as the first node when LOC=NULL
1. [OVERFLOW ?] If AVAIL==NULL then write OVERFLOW and EXIT
2. [Remove first node from AVAIL list]
3. Set INFO[NEW]=ITEM[copies new data into new node]
4. If LOC==NULL then :[insert as first node] Set LINK[NEW]=Start and START=NEW
else [insert after node with location LOC] Set LINK[NEW]=LINK[LOC] and LINK[LOC]=NEW
end of if structure
5. exit
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SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current
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SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current
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SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current
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SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current
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134
SINGLE LINKED LIST DELETION
DELETE(INFO, LINK,START, AVAIL, ITEM) This algorithm deletes from a linked list the first node N which contains the given ITEM of information
1. [Use procedure FIND given after this algorithm] FIND(INFO, LINK, START, ITEM, LOC, LOCP)
2. If LOC==NULL then: Write ITEM not in list and exit
3. If LOCP==NUL then Set START=LINK[START]
else : Set LINK[LOCP]=LINK[LOC]
3. {Return deleted node to the AVAIL list] Set LINK[LOC]=AVAIL and AVAIL=LOC
4. Exit
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135
SINGLE LINKED LIST DELETION
FINDB(INFO, START, ITEM, LOC, LOCP) This procedure finds the location LOC of first node N which contains ITEM and the location LOVP of the node preceding N. if ITEM does not appear in the list then the procedure sets LOC=NULL and if ITEM appears in the first node then it sets LOCP=NULL
1. [list empty?] if START==NULL then Set LOC =NULL and LOCP==NULL and return
2. [ITEM in first node ] if INFO[START]==ITEM then Set LOC=START and LOCP=NULL and return
3. Set SAVE=START and PRT=LINK[START]
4. Repeat steps 5 and 6 while PTR!=NULL
5. If INFO[PTR]==ITEM then Set LOC=PTR and LOCP=NULL
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SINGLE LINKED LIST DELETION
6. Set SAVE=PTR and PTR=LINK[PTR]
7. Set LOC=NULL
8. Return
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SINGLE LINKED LIST DELETION
6 X532start
LOCP LOC
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SINGLE LINKED LIST DELETION
6 X532start
LOCP LOC
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SINGLE LINKED LIST DELETION
6 X532start
LOCP LOC
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140
Double Linked Lists
Consists of nodes with two linked references one points to the previous and other to the next node
Maximises the needs of list traversals
Compared to single list inserting and deleting nodes is a bit slower as both the links had to be updated
It requires the extra storage space for the second list
A B Cstart
Tail
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141
Insertion Double Linked Lists
INSERTWL(INFO,FORW, BACK, STACK,AVAIL,LOCA,LOCB, ITEM)
1. [OVERFLOW?] If AVAIL==NULL then write OVER FLOW and exit
2. [Remove node from AVAIL list and copy new data into node] Set NEW=AVIAL, AVIAL=FORW[AVAIL] INFO[NEW]=ITEM
3. [Insert node into list] Set FORW[LOCA]=NEW, FORW[NEW]=LOCB BACK[LOCB]=NEW BACK[NEW]=LOCA
4. Exit
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Insertion Double Linked Lists
3 4 7start
TailNode A Node B
LOC A LOC B
7NEW
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Insertion Double Linked Lists
3 4 7start
TailNode A Node B
LOC A LOC B
7NEW
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Insertion Double Linked Lists
3 4 7start
TailNode A Node B
LOC A LOC B
7NEW
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145
Insertion Double Linked Lists
3 4 7start
TailNode A Node B
LOC A LOC B
7NEW
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146
Insertion Double Linked Lists
3 4 7start
TailNode A Node B
LOC A LOC B
7NEW
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147
Deletion Double Linked Lists
DELTWL(INFO, FORW, BACK, START, AVAIL, LOC)
1. [Delete node]
Set FORE[BACK[LOC]]=FORW[LOC]
Set BACK[FORW[LOC]]=BACK[LOC]
2. [Return node to AVAIL list]
Set FORW[LOC]=AVAIL and AVAIL=LOC
3. Exit
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Deletion Double Linked Lists
A B Cstart
Tail
LOC
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149
Deletion Double Linked Lists
A B Cstart
Tail
LOC
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150
Deletion Double Linked Lists
A B Cstart
Tail
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151
Deletion Double Linked Lists
A B Cstart
Tail