Data structure--Dynamic linked list (C + +)

Define a node:

#include <iostream>using namespace Std;typedef int t;struct node{t data; Node* Next; Node (const t& D):d Ata (d), next (NULL) {}operator T () {return data;}}; int main () {Node A (ten), B (+), cout << "a=" << a << ", b=" << b << endl;return 0;}

The above operator overloads first strongly convert the type A to the T type (that is, int), and ToString in Java is actually similar to a strongly converted string type.

Output a simple linked list

#include <iostream>using namespace Std;typedef int t;struct node{    T data;    Node* Next;    Node (const t& D):d Ata (d), next (NULL) {}    operator T () {        return data;    }   }; int main () {    Node A (Ten), B (+), C (+), D (+), E (+);    A.next = &b;     B.next = &c;     C.next = &d;     Node *p = &a;     while (P! = NULL) {        cout << *p << ';        p = p->next;    }    cout << Endl;    return 0;}

Add an element to the list

#include <iostream>using namespace Std;typedef int t;struct node{t data; Node* Next; Node (const t& D):d Ata (d), next (NULL) {}operator T () {return data;}};/ /Output list void Showlist (node* p) {while (P! = NULL) {cout << *p << ';p = p->next;} cout << Endl;} int main () {Node A (ten), B (+), C (+), D (+), E (+), A.next = &b;b.next = &c;c.next = &d;showlist (&a);//Add a section Point node* & p = b.next;//The alias of the b.next pointer e.next = P;p = &e;showlist (&a);//Add another node node* k = new node (70); node*& r = C.next;k->next = R;r = K;return 0;}

A list of C + + implementations is as follows:

#include <iostream>using namespace Std;typedef int t;class list{struct node{t data; Node * NEXT;//T () 0 initializes node (const t& d=t ()):d ATA (d), next (0) {}}; Node * HEAD; Head pointer int len;public:list (): Head (NULL), Len (0) {}//is inserted anywhere//1, a pointer to that location is found in the list PN//2, the next member of the new node and the PN point to the same place//3, Then let PN point to the new node void Insert (const t&d, int pos) {node*& pn = getptr (POS); node* p = new Node (d);p->next = PN;PN = p;len++;} Returns the list length int size () Const{return Len; End plug void push_back (const t& D) {insert (d, size ());} Find a pointer to a specified position in the linked list node*& getptr (int pos) {if (pos<0 | | pos>size ()) pos = 0;if (pos==0) return head; node* p = head;for (int i=1; i<pos; i++) p = P->next;return P->next;} Pre-inserted void Push_front (const t& D) {insert (d, 0);} Traverse void Travel () const{node* p = head;while (p!=null) {cout << p->data << ';p = p->next;} cout << Endl;} empty void Clear () {while (head!=null) {Node * p = head->next;delete head;head = p;} len = 0;} ~list () {clear ();} Delete by location//1, find the pointer to that position in the list//2, save the pointer to a copy//3, let that fingerPin points to the next node//4, releasing the pointer's dynamic memory void erase (int pos) {if (pos<0 | | pos>=size ()) return; Node *& pn = getptr (POS); Node * p = PN;PN = Pn->next;delete P;--len;} Find the position based on the element int find (const t& d) Const{int pos = 0; node* p = head;while (p) {if (P->data==d) return pos;p = P->next;++pos;} return-1;} remove void Remove (const t& d) {int Pos;while (pos = find (d))! =-1) Erase (POS) according to the element;}}; int main () {List l;l.push_front (5); L.push_front (8); L.push_front (20);//Insert 9l.insert (9, 2) in 2nd position; L.travel (); return 0;}

As can be seen, if we want to insert a node, we need to find a pointer to that position (or the previous node), such as the P->next pointer is what we need to find. Deleting a node also requires finding a pointer to that node.

Data structure--Dynamic linked list (C + +)