Data Structure-Dynamic Linked List (C ++)

Data Structure-Dynamic Linked List (C ++)

Define a node:

#include 
 
  using namespace std;typedef int T;struct Node{T data;Node* next;Node(const T& d):data(d), next(NULL){}operator T(){return data;}};int main(){Node a(10), b(20);cout << "a=" << a << ", b=" << b << endl;return 0;}
 

The preceding operator is overloaded. First, convert type a to type T (that is, int). in Java, the toString type is actually strongly converted to the string type.

Output a simple linked list

#include 
 
  using namespace std;typedef int T;struct Node{    T data;    Node* next;    Node(const T& d):data(d), next(NULL){}    operator T(){        return data;    }   };int main(){    Node a(10), b(20), c(30), d(40), e(50);    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 linked list

# Include
 
  
Using namespace std; typedef int T; struct Node {T data; Node * next; Node (const T & d): data (d), next (NULL) {} operator T () {return data ;}; // output linked table void showlist (Node * p) {while (p! = NULL) {cout <* p <''; p = p-> next;} cout <endl ;}int main () {Node a (10 ), B (20), c (30), d (40), e (50);. next = & B; B. next = & c; c. next = & d; showlist (& a); // Add a Node * & p = B. next; // obtain B. next pointer alias e. next = p; p = & e; showlist (& a); // Add another Node * k = new Node (70); Node * & r = c. next; k-> next = r; r = k; return 0 ;}
 

A linked list implemented by C ++ is as follows:

# Include
 
  
Using namespace std; typedef int T; class List {struct Node {T data; Node * next; // T () Zero initialization Node (const T & d = T ()): data (d), next (0) {}}; Node * head; // head pointer int len; public: List (): head (NULL), len (0) {} // insert to any position. // 1. Locate the pointer pn pointing to that position in the linked list. // 2. Point the next member and pn of the new node to the same place // 3. 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 ++ ;} // return the length of the linked list int size () const {return len;} // insert void push_back (const T & d) {insert (d, size ());} // find the Node * & getptr (int pos) {if (pos <0 | pos> size () pos = 0; if (pos = 0) return head; Node * p = head; for (int I = 1; I
  
   
Next; return p-> next;} // insert 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 ;}// clear void clear () {while (head! = NULL) {Node * p = head-> next; delete head; head = p;} len = 0 ;}~ List () {clear ();} // Delete by position // 1. Find the pointer pointing to that position in the linked list // 2. Save the pointer another // 3. Point the pointer to the next node // 4. Release the dynamic memory void erase (int pos) of the pointer) {if (pos <0 | pos> = size () return; Node * & pn = getptr (pos); Node * p = pn; pn = pn-> next; delete p; -- len;} // locate int find (const T & d) const {int pos = 0; Node * p = head; while (p) based on the element) {if (p-> data = d) return pos; p = p-> next; ++ pos;} return-1 ;} // Delete void remove (const T & d) {int pos; while (pos = find (d) based on the element ))! =-1) erase (pos) ;}}; int main () {List l; l. push_front (5); l. push_front (8); l. push_front (20); // insert 9l in 2nd locations. insert (9, 2); l. travel (); return 0 ;}
  
 

We can see that if we want to insert a node, we need to find the pointer pointing to this position (or the previous node). For example, the p-> next pointer is what we need to find. To delete a node, you must find a pointer to the node.