C - Count nodes in the Linked List
Counting nodes in a linked list is very useful while working on it. It requires creating a temp node pointing to the head of the list and a variable called i with initial value 0. If the temp node is not null, increase i by 1 and move to the next node using temp next. Repeat the process till the temp node becomes null. The final value of i will be the total number of nodes in the linked list.
The function countNodes is created for this purpose. It is a 4-step process.
int countNodes(struct Node* head_ref) { //1. create a temp node pointing to head struct Node* temp = head_ref; //2. create a variable to count nodes int i = 0; //3. if the temp node is not null increase // i by 1 and move to the next node, repeat // the process till the temp becomes null while (temp != NULL) { i++; temp = temp->next; } //4. return the count return i; }
The below is a complete program that uses above discussed concept of counting the total number of nodes of a linked list.
#include <stdio.h> #include <stdlib.h> //node structure struct Node { int data; struct Node* next; }; //Add new element at the end of the list void push_back(struct Node** head_ref, int newElement) { struct Node *newNode, *temp; newNode = (struct Node*)malloc(sizeof(struct Node)); newNode->data = newElement; newNode->next = NULL; if(*head_ref == NULL) { *head_ref = newNode; } else { temp = *head_ref; while(temp->next != NULL) { temp = temp->next; } temp->next = newNode; } } //count nodes in the list int countNodes(struct Node* head_ref) { struct Node* temp = head_ref; int i = 0; while (temp != NULL) { i++; temp = temp->next; } return i; } //display the content of the list void PrintList(struct Node* head_ref) { struct Node* temp = head_ref; if(head_ref != NULL) { printf("The list contains: "); while (temp != NULL) { printf("%i ",temp->data); temp = temp->next; } printf("\n"); } else { printf("The list is empty.\n"); } } // test the code int main() { struct Node* MyList = NULL; //Add four elements in the list. push_back(&MyList, 10); push_back(&MyList, 20); push_back(&MyList, 30); push_back(&MyList, 40); //Display the content of the list. PrintList(MyList); //number of nodes in the list printf("No. of nodes: %i\n",countNodes(MyList)); return 0; }
The above code will give the following output:
The list contains: 10 20 30 40 No. of nodes: 4