Mục lục bài viết
Update: 2021-12-06 18:02:09,Bạn Cần tương hỗ về P and q are pointers to a node of the linked list. Quý quý khách trọn vẹn có thể lại phản hồi ở cuối bài để Admin được tương hỗ.
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A linked list is a basic data structure where each item contains the information that we need to get to the next item.
The main advantage of linked lists over arrays is that the links provide us with the capability to rearrange the item efficiently. This flexibility is gained at the expense of quick access to any arbitrary item in the list, because the only way to access to an item in the list is to follow links from the beginning.
The following examples are for the linked list. Inside each example, we have several operations:
Also, there is another set of linked list quiz.
#include
using namespace std;
struct Node
int data;
Node* next;
;
// only for the 1st Node
void initNode(struct Node *head,int n)
head->data = n;
head->next =NULL;
// apending
void addNode(struct Node *head, int n)
Node *newNode = new Node;
newNode->data = n;
newNode->next = NULL;
Node *cur = head;
while(cur)
if(cur->next == NULL)
cur->next = newNode;
return;
cur = cur->next;
void insertFront(struct Node **head, int n)
Node *newNode = new Node;
newNode->data = n;
newNode->next = *head;
*head = newNode;
struct Node *searchNode(struct Node *head, int n)
Node *cur = head;
while(cur)
if(cur->data == n) return cur;
cur = cur->next;
cout << "No Node " << n <next;
delete ptrDel;
return true;
while(cur)
if(cur->next == ptrDel)
cur->next = ptrDel->next;
delete ptrDel;
return true;
cur = cur->next;
return false;
/* reverse the list */
struct Node* reverse(struct Node** head)
Node *parent = *head;
Node *me = parent->next;
Node *child = me->next;
/* make parent as tail */
parent->next = NULL;
while(child)
me->next = parent;
parent = me;
me = child;
child = child->next;
me->next = parent;
*head = me;
return *head;
/* Creating a copy of a linked list */
void copyLinkedList(struct Node *node, struct Node **pNew)
if(node != NULL)
*pNew = new Node;
(*pNew)->data = node->data;
(*pNew)->next = NULL;
copyLinkedList(node->next, &((*pNew)->next));
/* Compare two linked list */
/* return value: same(1), different(0) */
int compareLinkedList(struct Node *node1, struct Node *node2)
static int flag;
/* both lists are NULL */
if(node1 == NULL && node2 == NULL)
flag = 1;
else
if(node1 == NULL
return flag;
void deleteLinkedList(struct Node **node)
struct Node *tmpNode;
while(*node)
tmpNode = *node;
*node = tmpNode->next;
delete tmpNode;
void display(struct Node *head)
Node *list = head;
while(list)
cout <data <next;
cout << endl;
cout << endl;
int main()
struct Node *newHead;
struct Node *head = new Node;
initNode(head,10);
display(head);
addNode(head,20);
display(head);
addNode(head,30);
display(head);
addNode(head,35);
display(head);
addNode(head,40);
display(head);
insertFront(&head;,5);
display(head);
int numDel = 5;
Node *ptrDelete = searchNode(head,numDel);
if(deleteNode(&head;,ptrDelete))
cout << "Node "<< numDel << " deleted!n";
display(head);
cout << "The list is reversedn";
reverse(&head;);
display(head);
cout << "The list is copiedn";
copyLinkedList(head,&newHead;);
display(newHead);
cout << "Comparing the two lists…n";
cout << "Are the two lists same?n";
if(compareLinkedList(head,newHead))
cout << "Yes, they are same!n";
else
cout << "No, they are different!n";
cout << endl;
numDel = 35;
ptrDelete = searchNode(newHead,numDel);
if(deleteNode(&newHead;,ptrDelete))
cout << "Node "<< numDel << " deleted!n";
cout << "The new list after the delete isn";
display(newHead);
cout << "Comparing the two lists again…n";
cout << "Are the two lists same?n";
if(compareLinkedList(head,newHead))
cout << "Yes, they are same!n";
else
cout << "No, they are different!n";
cout << endl;
cout << "Deleting the copied listn";
deleteLinkedList(&newHead;);
display(newHead);
return 0;
Output from the run:
10
10 20
10 20 30
10 20 30 35
10 20 30 35 40
5 10 20 30 35 40
Node 5 deleted!
10 20 30 35 40
The list is reversed
40 35 30 20 10
The list is copied
40 35 30 20 10
Comparing the two lists…
Are the two lists same?
Yes, they are same!
Node 35 deleted!
The new list after the delete is
40 30 20 10
Comparing the two lists again…
Are the two lists same?
No, they are different!
Deleting the copied list
#include
using namespace std;
struct node
int data;
struct node * next;
;
node *head = NULL;
// returning the pointer to the element
// whose data is less than or equal to input data
struct node *searchNode(int n)
if(head == NULL) return head;
node *cur = head;
node *prev = head;
while(cur)
if(cur->data == n) return cur;
if(cur->data > n) return prev;
prev = cur;
cur = cur->next;
// returning the pointer to the element
// whose data is equal to input data
struct node *searchNode2(int n)
if(head == NULL) return head;
node *cur = head;
node *prev = head;
while(cur)
if(cur->data == n) return cur;
prev = cur;
cur = cur->next;
return cur;
void addNode(int n)
node *newNode = new node;
newNode->data = n;
newNode->next = NULL;
if(head == NULL)
head = newNode;
return;
node *cur = head;
while(cur)
if(cur->next == NULL)
cur->next = newNode;
return;
cur = cur->next;
void insertNode(int n)
node *ptr = searchNode(n);
node *newNode = new node;
newNode->data = n;
node *cur = head;
while(cur)
if(cur == ptr )
newNode->next = cur->next;
cur->next = newNode;
return;
cur = cur->next;
void deleteNode(int n)
node *ptr = searchNode(n);
if(ptr == NULL)
cout << "No node with data = " << n <next;
return;
node *cur = head;
node *prev = head;
while(cur)
if(cur == ptr)
prev->next = cur->next;
return;
prev = cur;
cur = cur->next;
void display()
struct node *list = head;
while(list)
cout <data <next;
cout <name, name );
ptr->id = id;
return ptr;
// adding to the end of list
void addNode( struct node *newnode )
// if there is no node, put it to head
if( head == NULL )
head = newnode;
tail = newnode;
// link in the new_node to the tail of the list
// then mark the next field as the end of the list
// adjust tail to point to the last node
tail->next = newnode;
newnode->next = NULL;
tail = newnode;
void insertNode( struct node *newnode )
struct node *temp, *prev;
if( head == NULL ) // if an empty list,
head = newnode; // set ‘head’ to it
tail = newnode;
head->next = NULL; // set end of list to NULL
return;
temp = head; // start at beginning of list
// while currentname name, newnode->name) next; // goto the next node in list
if( temp == NULL ) // don’t go past end of list
break;
// set previous node before we insert
// first check to see if it’s inserting
if( temp == head ) // before the first node
newnode->next = head; // link next field to original list
head = newnode; // head adjusted to new node
else // it’s not the first node
prev = head; // start of the list,
while( prev->next != temp ) // will cycle to node before temp
prev = prev->next;
prev->next = newnode; // insert node between prev and next
newnode->next = temp;
if( tail == prev ) // if the new node is inserted at the
tail = newnode; // end of the list the adjust ‘end’
struct node * searchName( struct node *ptr, char *name )
while( strcmp( name, ptr->name ) != 0 )
ptr = ptr->next;
if( ptr == NULL )
break;
return ptr;
struct node* searchId(struct node* ptr, int id)
while( id != ptr->id )
ptr = ptr->next;
if( ptr == NULL )
break;
return ptr;
void reverse()
// we need at least two nodes for the reverse to have any effect
if(head == NULL
void deleteNode( struct node *ptr )
struct node *temp, *prev;
temp = ptr; // node to be deleted
prev = head; // start of the list, will cycle to node before temp
if( temp == prev ) // deleting first node?
head = head->next; // moves head to next node
if( tail == temp ) // is it end, only one node?
tail = tail->next; // adjust end as well
delete temp ; // không lấy phí up space
else // if not the first node, then
while( prev->next != temp ) // move prev to the node before
prev = prev->next; // the one to be deleted
prev->next = temp->next; // link previous node to next
if( tail == temp ) // if this was the end node,
tail = prev; // then reset the end pointer
delete temp; // không lấy phí up space
void deleteList( struct node *ptr )
struct node *temp;
if( head == NULL ) return; // don’t try to delete an empty list
if( ptr == head ) // if we are deleting the entire list
head = NULL; // then reset head and
tail = NULL; // end to empty
else
temp = head; // if it’s not the entire list, readjust end
while( temp->next != ptr ) // locate previous node to ptr
temp = temp->next;
tail = temp; // set end to node before ptr
while( ptr != NULL ) // while there are still nodes to delete
temp = ptr->next; // record address of next node
delete ptr; // không lấy phí this node
ptr = temp; // point to next node to be deleted
void displayNode( struct node *ptr )
cout <id << ": " <name <next;
#include
using namespace std;
int main()
char* name;
int id, ch = 1;
struct node *ptr;
// add
ptr = initNode( “s1”, 1 );
addNode(ptr);
ptr = initNode( “s2”, 2 );
addNode(ptr);
ptr = initNode( “s3”, 3 );
addNode(ptr);
ptr = initNode( “s4”, 4 );
addNode(ptr);
ptr = initNode( “s5”, 5 );
addNode(ptr);
displayList(head);
// delete
name = “s2”;
ptr = searchName(head, name );
if( ptr == NULL )
cout << "nName: " << name << " not found" << endl;
else
cout << "nDeleting a node … ";
displayNode(ptr);
deleteNode( ptr );
displayList(head);
// insert
name = "s2";
id = 2;
ptr = initNode( name, id );
insertNode( ptr );
cout << "nInserting a node … ";
displayNode(ptr);
displayList(head);
// reverse
cout << "nReversing the list … n";
reverse();
displayList(head);
// delete
cout <next;
delete temp;
temp = current;
struct list::node* list::initNode(string s, int i)
struct node *ptr = new node;
// error? then just return
if( ptr == NULL )
return static_cast(NULL);
// assign it
// then return pointer to ne node
else
ptr->name = s ;
ptr->id = i;
return ptr;
// adding to the end of list
void list::addNode( struct node *newNode )
// if there is no node, put it to head
if( head == NULL )
head = newNode;
tail = newNode;
// link in the new_node to the tail of the list
// then mark the next field as the end of the list
// adjust tail to point to the last node
tail->next = newNode;
newNode->next = NULL;
tail = newNode;
void list::insertNode( struct node *newnode )
struct node *temp, *prev;
if( head == NULL ) // if an empty list,
head = newnode; // set ‘head’ to it
tail = newnode;
head->next = NULL; // set end of list to NULL
return;
temp = head; // start at beginning of list
// while currentname name name) // to be inserted then
temp = temp->next; // goto the next node in list
if( temp == NULL ) // don’t go past end of list
break;
// set previous node before we insert
// first check to see if it’s inserting
if( temp == head ) // before the first node
newnode->next = head; // link next field to original list
head = newnode; // head adjusted to new node
else // it’s not the first node
prev = head; // start of the list,
while( prev->next != temp )
prev = prev->next; // will cycle to node before temp
prev->next = newnode; // insert node between prev and next
newnode->next = temp;
if( tail == prev ) // if the new node is inserted at the
tail = newnode; // end of the list the adjust ‘end’
struct list::node* list::searchName(struct node* ptr, string name)
while( name != ptr->name )
ptr = ptr->next;
if( ptr == NULL )
break;
return ptr;
struct list::node* list::searchId(struct node* ptr, int id)
while( id != ptr->id )
ptr = ptr->next;
if( ptr == NULL )
break;
return ptr;
void list::reverse()
// we need at least two nodes for the reverse to have any effect
if(head == NULL
void list::deleteNode( struct list::node *ptr )
struct node *temp, *prev;
temp = ptr; // node to be deleted
prev = head; // start of the list, will cycle to node before temp
if( temp == prev ) // deleting first node?
head = head->next; // moves head to next node
if( tail == temp ) // is it end, only one node?
tail = tail->next; // adjust end as well
delete temp ; // không lấy phí up space
else // if not the first node, then
while( prev->next != temp ) // move prev to the node before
prev = prev->next; // the one to be deleted
prev->next = temp->next; // link previous node to next
if( tail == temp ) // if this was the end node,
tail = prev; // then reset the end pointer
delete temp; // không lấy phí up space
void list::deleteList( struct node *ptr )
struct node *temp;
if( head == NULL ) return; // don’t try to delete an empty list
if( ptr == head ) // if we are deleting the entire list
head = NULL; // then reset head and
tail = NULL; // end to empty
else
temp = head; // if it’s not the entire list, readjust end
while( temp->next != ptr ) // locate previous node to ptr
temp = temp->next;
tail = temp; // set end to node before ptr
while( ptr != NULL ) // whilst there are still nodes to delete
temp = ptr->next; // record address of next node
delete ptr; // không lấy phí this node
ptr = temp; // point to next node to be deleted
void list::displayNode( struct list::node *ptr ) const
cout <id << ": " <name << endl;
void list::displayList( struct list::node *ptr ) const
if(!ptr) cout << "Nothing to display" <next;
int main()
int id;
string name;
list myList;
list::node* ptr;
// add
ptr = myList.initNode( “s1”, 1 );
myList.addNode(ptr);
ptr = myList.initNode( “s2”, 2 );
myList.addNode(ptr);
ptr = myList.initNode( “s3”, 3 );
myList.addNode(ptr);
ptr = myList.initNode( “s4”, 4 );
myList.addNode(ptr);
ptr = myList.initNode( “s5”, 5 );
myList.addNode(ptr);
myList.displayList(myList.head);
// delete
name = “s2”;
ptr = myList.searchName( myList.head, name );
if( ptr == NULL )
cout << "nName: " << name << " not found" << endl;
else
cout << "nDeleting a node … ";
myList.displayNode(ptr);
myList.deleteNode( ptr );
myList.displayList(myList.head);
// insert
name = "s2";
id = 2;
ptr = myList.initNode( name, id );
myList.insertNode( ptr );
cout << "nInserting a node … ";
myList.displayNode(ptr);
myList.displayList(myList.head);
// reverse
cout << "nReversing the list … n";
myList.reverse();
myList.displayList(myList.head);
// delete
cout << "nIn the end, deleting the list … n";
myList.deleteList(myList.head);
myList.displayList(myList.head);
return 0;
Example 5A – Detecting Circular (Loop) Linked List
/* This code has the following */
/*
1. Adding Nodes
2. Function returning the size of the list
3. Making the list circular (loop)
4. Detecting circular list
5. Recursive printing
*/
#include
using namespace std;
struct Node
int data;
Node * next;
;
Node *root = 0;
void addNode(int n)
if(root==0)
root = new Node;
root->data = n;
root->next = 0;
return;
Node *cur = root;
while(cur)
if(cur->next == 0)
Node *ptr = new Node;
ptr -> data = n;
ptr -> next = 0;
cur -> next = ptr;
return;
cur = cur->next;
void makeCircular()
!root->next) return;
Node *cur = root;
while(cur)
if(cur->next == 0)
cur->next = root;
return;
cur = cur->next;
int sizeOfList()
Node *cur = root;
int size = 0;
while(cur)
size++;
if(cur->next == 0)
return size;
cur = cur->next;
return size;
bool detectCircular()
void printRecursive(Node *ptr)
if(!ptr)
cout << endl;
return;
cout <data <next);
int main(int argc, char **argv)
addNode(10);addNode(20);addNode(30);addNode(40);addNode(50);
addNode(60);addNode(70);addNode(80);addNode(90);addNode(100);
printRecursive(root);
cout << "size of list = " << sizeOfList() << endl;
makeCircular();
if(detectCircular()) cout <<"Circularn";
else cout << "Normaln";
Output from the run:
10 20 30 40 50 60 70 80 90 100
size of list = 10
20,30
30,50
40,70
50,90
60,10
70,30
80,50
90,70
100,90
10,10
Circular
Example 5B – Detecting Circular (Loop) Linked List (Generic Class with Template)
#include
#include
using namespace std;
template
class LinkedList
private:
struct node
T data;
node * next;
*head;
public:
LinkedList();
~LinkedList();
void add(T d);
void remove(T d);
void clear();
void makeCircular();
bool isCircular();
void display(const char* s);
;
template
LinkedList::LinkedList()
head = NULL;
template
LinkedList::~LinkedList()
node *p., *q;
p. = head;
if(p. == NULL) return;
while(p.)
q = p.->next;
delete p.;
p. = q;
template
void LinkedList::add(T d)
node *p., *q;
if(head == NULL)
head = new node;
head->data = d;
head->next = NULL;
return;
p. = head;
while(p.->next != NULL)
p. = p.->next;
q = new node;
q->data = d;
q->next = NULL;
p.->next = q;
template
void LinkedList::remove(T d)
node *p., *q;
if(head == NULL) return;
p. = head;
while(p.)
if(p.->data == d)
q->next = p.->next;
delete p.;
return;
q = p.;
p. = p.->next;
template
void LinkedList::clear()
node *p., *q;
if(head == NULL) return;
p. = head;
while(p.)
q = p.->next;
delete p.;
if(q != head)
head = NULL;
return;
p. = q;
template
void LinkedList::makeCircular()
node *p.;
if(head == NULL
template
bool LinkedList::isCircular()
node *p., *pp;
if(head == NULL
template
void LinkedList::display(const char* s)
node *p.;
if(head == NULL) return;
p. = head;
while(p.)
if(s == “string”)
cout <
data << endl;
else
cout <
data <next;
if(p. != NULL)
if(p. == head) return;
if(s == “integer”) cout << endl;
int main()
LinkedList sList;
sList.add("Wolfgang Amadeus Mozart");
sList.add("Franz Peter Schubert");
sList.add("Pyotr Ilyich Tchaikovsky");
sList.add("Clude-Achille Debussy");
sList.add("Camille Saint-Saens");
sList.display("string");
sList.clear();
LinkedList iList;
iList.add(40);
iList.add(50);
iList.add(60);
iList.add(70);
iList.add(80);
iList.add(90);
iList.add(100);
iList.add(10);
iList.add(20);
iList.add(30);
iList.display("integer");
/* link last element to the first */
iList.makeCircular();
if(iList.isCircular()) cout <<"This is circular listn";
iList.display("integer");
iList.clear();
cout << "ndisplay after clear()n";
iList.display("integer");
return 0;
Output from the run is:
Wolfgang Amadeus Mozart
Franz Peter Schubert
Pyotr Ilyich Tchaikovsky
Clude-Achille Debussy
Camille Saint-Saens
40 50 60 70 80 90 100 10 20 30
This is circular list
40 50 60 70 80 90 100 10 20 30
display after clear()
Example 6 – Stack Using Linked List
This stack is using linked list as its data structure.
/* Stack operations */
#include
using namespace std;
typedef struct Element
void *data;
struct Element *next;
Element;
bool push(Element **top, void *data)
Element *elem = new Element;
if(!elem) return false;
elem->data = data;
elem->next = *top;
*top = elem;
return true;
bool createStack(Element **top)
*top = NULL;
return true;
bool pop(Element **top, void **data )
Element *elem;
if( !(elem = *top) ) return false;
*data = elem->data;
*top = elem->next;
delete elem;
return true;
bool deleteStack(Element **top)
Element *elem;
while(*top)
elem = (*top)->next;
delete *top;
*top = elem;
return true;
void printStack(Element *top)
if(top==NULL)
cout << "Empty!" << endl;
Element *cur = top;
while(cur)
cout <data)) <next;
cout << endl << endl;
int main()
Element *head;
createStack(&head;);
int n1 = 10, n2 = 20, n3 = 30, n4 = 40, n5 = 50;
push(&head;, &n1;);
push(&head;, &n2;);
push(&head;, &n3;);
push(&head;, &n4;);
push(&head;, &n5;);
printStack(head);
void * n;
pop(&head;, &n;);
cout << "popped " << *(static_cast(n)) << endl;
pop(&head;, &n;);
cout << "popped " << *(static_cast(n)) << endl;
cout << endl;
printStack(head);
cout << "deleting stack…" << endl;
deleteStack(&head;);
printStack(head);
Output:
50 40 30 20 10
popped 50
popped 40
30 20 10
deleting stack…
Empty!
Example 7 – Stack Class Using Linked List
#include
using namespace std;
class Stack
public:
Stack();
~Stack();
void push(int);
int pop();
int peek();
friend void print(Stack&);
private:
typedef struct node
node *next;
int data;
NODE;
NODE *top;
;
Stack::Stack()
top = NULL;
Stack::~Stack()
while(top)
NODE *tmp = top;
top = top->next;
delete tmp;
void Stack::push(int n)
NODE *ptr = new NODE;
ptr->next = top;
ptr->data = n;
top = ptr;
int Stack::pop()
NODE *tmp = top;
int val = top->data;
top = top->next;
delete tmp;
return val;
int Stack::peek()
return top->data;
void print(Stack &s;)
Stack::NODE *cur = s.top;
while(cur)
cout <data <next;
cout <push(10);
st->push(20);
st->push(30);
st->push(40);
st->push(50);
print(*st);
st->pop();
st->pop();
print(*st);
cout << "current top=" <peek();
return 0;
Output:
50 40 30 20 10
30 20 10
current top=30
Example 7B – Stack Class Using Linked List
The code below is almost the same as the code in Example 6 except it’s using Stack class.
The code below is almost the same as the code in Example 7 except it’s using void* for the data type.
#include
using namespace std;
class Stack
public:
Stack();
~Stack();
void push(void *data);
void *pop();
void print();
protected:
typedef struct Element
struct Element *next;
void *data;
Element;
Element *top;
;
Stack::Stack()
top = NULL;
Stack::~Stack()
while(top)
Element *elm = top->next;
delete top;
top = elm;
void Stack::push(void *data)
Element *elm = new Element;
elm->data = data;
elm->next = top;
top = elm;
void *Stack::pop()
void *data;
if(top == NULL) return data;
data = top->data;
Element *elm = top;
top = elm->next;
delete elm;
return data;
void Stack::print()
Element *elm = top;
while(elm)
cout <data)) <next;
cout <push(&n1;);
st->push(&n2;);
st->push(&n3;);
st->push(&n4;);
st->push(&n5;);
st->print();
cout <pop()))<< " popedn";
cout <pop()))<print();
cout << endl;
Output:
50 40 30 20 10
50 poped
40 poped
30 20 10
Example 7C – Stack Class Using Linked List with Query for Minimum Value
This stack class can return its minimum element. All operations (push(), pop(), and peekMin()) are O(1) not O(n). Usual approach for query would be traverse each element to get the minimum, and it will ends up with O(n) complexity. So, to have constant time operation, we need keep track of the minimum. We could have a global minimum value, however, it has a problem when the stack with the value popped. Then, we need to update the global min value which may take O(n) operation.
In the code below, each stack gets its own min-value at the time when it was pushed by comparing the minimum of the previous top stack. When the top stack popped, the code checks if the top stack’s minimum value is the global min.
In summary, this code will get the stacks minimum value by peeking the top only, using peekMin() which returns the min-value for the top stack.
#include
using namespace std;
#define MIN(a,b) (a next;
delete tmp;
void Stack::push(int n)
NODE *ptr = new NODE;
ptr->next = top;
ptr->data = n;
// currently empty (top is NULL)
if(top == NULL)
ptr->min = n;
else
ptr->min = MIN(n, top->min);
top = ptr;
int Stack::pop()
NODE *tmp = top;
int val = top->data;
top = top->next;
delete tmp;
cout << "pop " << val <data;
int Stack::peekMin()
return top->min;
void print(Stack &s;)
Stack::NODE *cur = s.top;
while(cur)
cout <data <next;
cout <push(40);
st->push(50);
st->push(20);
st->push(10);
st->push(30);
print(*st);
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() << endl;
print(*st);
cout << "current top=" <peek();
return 0;
Output:
30 10 20 50 40
minimum = 10
pop 30
minimum = 10
pop 10
minimum = 20
pop 20
minimum = 40
pop 50
minimum = 40
40
current top=40
Example 7D – Stack Class Using Linked List with Query for Minimum Value (additional stack)
Even though the code in Example 7C can keep track of the minimum of the stack, it is obvious that the code is wasting resources. Suppose, we push the element with the minimum first: 10, 20, 30, 40 … with increasing order. In that case, the every stack element has additional thành viên for minimum value of 10. However, if we have separate stack just for minimum, we end up having only one stack because we do not save the values which are greater than the min-value.
Here is the code using separate stack (StackWithMin class) just for the minimum value.
#include
using namespace std;
typedef struct node
node *next;
int data;
NODE;
class Stack
public:
Stack();
virtual ~Stack();
virtual void push(int);
virtual int pop();
virtual int peekMin();
int peek();
friend void print(Stack&);
private:
bool empty();
NODE *top;
;
Stack::Stack()
top = NULL;
Stack::~Stack()
while(top)
NODE *tmp = top;
top = top->next;
delete tmp;
void Stack::push(int n)
NODE *ptr = new NODE;
ptr->next = top;
ptr->data = n;
top = ptr;
int Stack::pop()
NODE *tmp = top;
if(!empty())
int val = top->data;
top = top->next;
delete tmp;
cout << "pop " << val << endl;
return val;
else
cout << "empty! " <data;
return -1;
bool Stack::empty()
if(top == NULL) return true;
return false;
int Stack::peekMin()
return -1;
void print(Stack &s;)
NODE *cur = s.top;
while(cur)
cout <data <next;
cout <next;
delete tmp;
void StackWithMin::push(int n)
if(top)
// push only if it’s smaller than the top min
if(n data)
NODE *ptr = new NODE;
ptr->next = top;
ptr->data = n;
top = ptr;
// if empty, just push the new element
else
NODE *ptr = new NODE;
ptr->next = top;
ptr->data = n;
top = ptr;
Stack::push(n);
int StackWithMin::pop()
int popped = Stack::pop();
if(empty())
cout << "empty min stack" <data)
NODE *tmp = top;
if(top->next)
top = top->next;
else
top = NULL;
delete tmp;
return popped;
int StackWithMin::peekMin()
if(!empty()) return top->data;
cout << "empty min stack!" <push(40);
st->push(50);
st->push(20);
st->push(10);
st->push(30);
print(*st);
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() <pop();
cout << "minimum = " <peekMin() << endl;
print(*st);
cout << "current top=" <peek();
return 0;
Output:
30 10 20 50 40
minimum = 10
pop 30
minimum = 10
pop 10
minimum = 20
pop 20
minimum = 40
pop 50
minimum = 40
pop 40
empty min stack!
minimum = -1
Example 8 Queue Struct : Using Linked List
#include
#include
struct node
int data;
node *next;
;
typedef struct node node_t;
node_t *head = NULL;
void push(int n)
node_t *newNode = (node_t *)malloc(sizeof(node_t));
newNode->data = n;
newNode->next = NULL;
if(head == NULL)
head = newNode;
return;
node_t *cur = head;
while(cur)
if(cur->next==NULL)
cur->next = newNode;
return;
cur = cur->next;
void pop()
if(head==NULL) return;
node_t *tmp = head;
head = head->next;
không lấy phí(tmp);
void display()
node_t *cur = head;
while(cur)
printf(“%3d”,cur->data);
cur = cur->next;
printf(“n”);
int main()
push(1);push(2);push(3);push(4);push(5);display();
pop();display();
pop();display();
pop();display();
pop();display();
pop();display();
return 0;
Output is:
1 2 3 4 5
2 3 4 5
3 4 5
4 5
5
Example 8B – Queue Class: Using Linked List
First one becomes head, so when it pops, head will be popped.
#include
using namespace std;
class Queue
public:
Queue();
~Queue();
void push(int);
int pop();
void print();
private:
typedef struct Node
Node *next;
int data;
NODE;
NODE* head;
;
Queue::Queue()
head = NULL;
Queue::~Queue()
if(head == NULL) return;
NODE *cur = head;
while(cur)
Node *ptr = cur;
cur = cur->next;
delete ptr;
void Queue::push(int n)
if(head == NULL)
head = new NODE;
head->data = n;
head->next = NULL;
return;
NODE *cur = head;
while(cur)
if(cur->next == NULL)
NODE *ptr = new NODE;
ptr->data = n;
ptr->next = NULL;
cur->next = ptr;
return;
cur = cur->next;
void Queue::print()
if(head==NULL) return;
Node *cur = head;
while(cur)
cout <data <next;
cout << endl;
int Queue::pop()
if(head == NULL)
cout << "empty estack!" <data;
if(head->next)
head = head->next;
// pop the last element (head)
else
delete tmp;
head = NULL;
cout << "pop: " << value <push(10);
que->push(20);
que->push(30);
que->push(40);
que->push(50);
que->print();
que->pop();que->print();
que->pop();que->print();
que->pop();que->print();
que->pop();que->print();
que->pop();que->print();
que->pop();que->print();
return 0;
Output:
10 20 30 40 50
pop: 10
20 30 40 50
pop: 20
30 40 50
pop: 30
40 50
pop: 40
50
pop: 50
Example 9 – Finding Intersection and Union of Two Linked List
The following code finds intersection/union of two linked list and puts it into a new linked list.
#include
using namespace std;
struct node
int data;
node *next;
;
void add(struct node **head, int n)
struct node *cur;
struct node *new_node = (struct node *)malloc(sizeof(struct node));
new_node->data = n;
new_node->next = NULL;
if(*head == NULL)
*head = new_node;
return;
cur = *head;
while(cur)
if(cur->next == NULL)
cur->next = new_node;
return;
cur = cur->next;
bool isDuplicate(struct node *head, int n)
struct node* cur = head;
while(cur)
if(cur->data == n) return true;
cur = cur->next;
return false;
struct node *getIntersection(struct node *headA, struct node *headB)
curB == NULL) return NULL;
while(curA)
while(curB)
if(curA->data == curB->data)
add(&result;, curA->data);
curB = curB->next;
curB = headB;
curA = curA->next;
return result;
struct node *getUnion(struct node *headA, struct node *headB)
struct node *cur;
struct node *result = NULL;
if(headA == NULL && headB == NULL) return NULL;
cur = headA;
while(cur)
add(&result;, cur->data);
cur = cur->next;
cur = headB;
while(cur)
/* check if the new data is already there */
if(!isDuplicate(result, cur->data))
add(&result;, cur->data);
cur = cur->next;
return result;
void display(struct node *head)
if(head == NULL) return;
struct node *cur = head;
while(cur)
cout <data <next;
cout << endl;
int main()
struct node *myListA = NULL;
struct node *myListB = NULL;
struct node *intersectionList = NULL;
struct node *unionList = NULL;
add(&myListA;,10);
add(&myListA;,20);
add(&myListA;,30);
add(&myListA;,40);
add(&myListA;,50);
add(&myListA;,60);
add(&myListA;,70);
cout << "List A: ";
display(myListA);
add(&myListB;,10);
add(&myListB;,30);
add(&myListB;,50);
add(&myListB;,70);
add(&myListB;,90);
add(&myListB;,110);
add(&myListB;,130);
cout << "List B: ";
display(myListB);
cout << "Intersection of A and B: ";
intersectionList = getIntersection(myListA, myListB);
display(intersectionList);
cout << "Union of A and B: ";
unionList = getUnion(myListA, myListB);
display(unionList);
return 0;
Output is:
List A: 10 20 30 40 50 60 70
List B: 10 30 50 70 90 110 130
Intersection of A and B: 10 30 50 70
Union of A and B: 10 20 30 40 50 60 70 90 110 130
Example 10 – Another Example of Generic Linked List
The following example is another example of generic use of linked list. It will be used as a base for doubly linked list later.
#include
using namespace std;
template
class List
struct Node
T data;
Node *next;
Node(T d, Node *n = 0):data(d), next(n)
;
Node *head;
public:
List(Node *h = 0):head(h)
~List();
void insert(Node *loc, T d);
void push_back(T d);
void push_front(T d);
T pop_back();
T pop_front();
void erase(Node *loc);
void display();
Node *search(T d);
;
// destructor
template
List::~List()
Node *tmp;
while(head)
tmp = head;
head = head->next;
delete tmp;
// insert d before loc
template
void List::insert(Node *loc, T d)
Node *new_node = new Node(d,0);
if(!head)
head = new_node;
return;
if(loc == head)
push_front(d);
return;
Node *cur = head;
while(cur->next)
if(cur->next == loc)
new_node->next = cur->next;
cur->next = new_node;
return ;
cur = cur->next;
template
void List::push_back(T d)
Node *new_node = new Node(d,0);
if(!head)
head = new_node;
return;
Node *cur = head;
while(cur)
if(!cur->next)
cur->next = new_node;
return;
cur = cur->next;
template
void List::push_front(T d)
Node *new_node = new Node(d,0);
if(!head)
head = new_node;
return;
new_node->next = head;
head = new_node;
return;
template
T List::pop_back()
Node *cur = head;
while(cur)
if(!cur->next)
T data (cur->data);
delete cur;
head = NULL;
return data;
else
if(!cur->next->next)
T data (cur->next->data);
cur->next = NULL;
delete cur->next;
return data;
cur = cur->next;
return NULL;
template
T List::pop_front()
if(!head) return NULL;
Node *tmp = head;
T data (head->data);
if(head->next)
head = head->next;
delete tmp;
return data;
delete tmp;
head = NULL;
return data;
template
void List::erase(Node *loc)
if(loc == head)
Node *tmp = head;
head = head->next;
delete tmp;
return;
Node *cur = head;
while(cur)
if(cur->next == loc)
cur->next = loc->next;
delete loc;
cur = cur->next;
template
typename List::Node* List::search(T d)
if(!head) return NULL;
Node* cur = head;
while(cur)
if(cur->data == d) return cur;
cur = cur->next;
return NULL;
template
void List::display()
if(!head) return;
Node *cur = head;
while(cur)
cout <data << " " <next;
cout << endl;
int main()
List *myList = new List(NULL);
cout <push_back(20);
myList->push_back(30);
myList->push_back(40);
myList->push_back(50);
myList->display();
cout <push_front(10);
myList->display();
cout <erase(myList->search(30));
myList->display();
cout <insert(myList->search(40),30);
myList->display();
cout << "pop_back()n";
cout <pop_back() <display();
cout << "pop_front()n";
cout <pop_front() <display();
return 0;
Output:
push_back() 20, 30 40, 50
20
30
40
50
push_front() 10
10
20
30
40
50
erase 30
10
20
40
50
insert 30 before 40
10
20
30
40
50
pop_back()
50 just back popped
10
20
30
40
pop_front()
10 just front popped
20
30
40
List of Linked List Examples of This Page
Also, there are set of linked list samples:
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