Question

How to create sorted linked list in C

Answer 1

#include <stdio.h>
#include <stdlib.h>
 
typedef struct Node {
    int x;
    struct Node *next;
} Node;
  
void add_first(Node **root, int value) {
    Node *new_node = malloc(sizeof(int));
    if (new_node == NULL) {
        puts("malloc error");
        exit(1);
    }
     
    new_node->x = value;
    new_node->next = *root;
     
    *root = new_node;
} 

void add_after(Node *node, int value) {
    Node* new_node = malloc(sizeof(Node));
     
    if (new_node == NULL) {
        puts("malloc error");
        exit(1);
    }
     
    new_node->x = value;
    new_node->next = node->next;
    node->next = new_node;
}

void add_sorted(Node **root, int value) {
    if (*root == NULL || (*root)->x >= value) {
        printf("add_first value %d\n", value);
        add_first(root, value);
        return;
    }
    
    Node *current = *root;
    while (current->next != NULL) {
        if (current->next->x >= value) {
            printf("current->x %d add_after value %d\n", current->x, value);
            add_after(current, value);
            return;
        }
        current = current->next;
    }
    printf("current->x %d add_afte value %d\n", current->x, value);
    add_after(current, value);
}
  
void add_node(Node **root, int value) {
    Node *new_node = malloc(sizeof(Node));
      
    if (new_node == NULL) {
        puts("malloc error");
        exit(1);
    }
      
    new_node->next = NULL;
    new_node->x = value;
      
    if (*root == NULL) {
        *root = new_node;
        return;
    }
      
    Node* current = *root;
    while (current->next != NULL) {
        current = current->next;
    }
    current->next = new_node;
}
  
void free_LinkedList(Node *root) {
    Node *next;
    while (root != NULL)  {  
        next = root->next;  
        free(root);  
        root = next;  
    }  
}
  
int main() {
    Node *root = NULL;
      
    add_sorted(&root, 7);
    add_sorted(&root, 2000);
    add_sorted(&root, 78);
    add_sorted(&root, 3);
    
    for (Node *current = root; current != NULL; current = current->next) {
        printf("%d\n", current->x);
    }

    free_LinkedList(root);
  
    return 0;
}
  
  
  
/*
run:
  
add_first value 7
current->x 7 add_afte value 2000
current->x 7 add_after value 78
add_first value 3
3
7
78
2000

*/

 

70+ SQL courses for beginners and professionals