


In the C program, translate the following content into Chinese: Program to find the nth node from the bottom of a linked list
Given n nodes, the task is to print the nth node at the end of the linked list. The program must not change the order of the nodes in the list, but should only print the nth node from the last node of the linked list.
Example
Input -: 10 20 30 40 50 60 N=3 Output -: 40
In the above example, starting from the first node, traverse to count-n nodes, that is, 10,20 30,40, 50,60, so the last Three nodes are 40.
Instead of traversing the entire list so efficiently one can follow -
- Get a temporary pointer to, say, a temp# of node type ##Set this temporary pointer to the first node head pointer pointed to
- Set the counter to the number of nodes in the list
- Move temp to temp → next until count -n
- Display temp → data
th Starting at position 10, the result up to 20 is in the 1st position, and the 30th position is in the second position. So with this approach, there is no need to iterate through the entire list until the end, which will save space and memory.
AlgorithmStart
Step 1 -> create structure of a node and temp, next and head as pointer to a structure node
struct node
int data
struct node *next, *head, *temp
End
Step 2 -> declare function to insert a node in a list
void insert(int val)
struct node* newnode = (struct node*)malloc(sizeof(struct node))
newnode->data = val
IF head= NULL
set head = newnode
set head->next = NULL
End
Else
Set temp=head
Loop While temp->next!=NULL
Set temp=temp->next
End
Set newnode->next=NULL
Set temp->next=newnode
End
Step 3 -> Declare a function to display list
void display()
IF head=NULL
Print no node
End
Else
Set temp=head
Loop While temp!=NULL
Print temp->data
Set temp=temp->next
End
End
Step 4 -> declare a function to find nth node from last of a linked list
void last(int n)
declare int product=1, i
Set temp=head
Loop For i=0 and i<count-n and i++
Set temp=temp->next
End
Print temp->data
Step 5 -> in main()
Create nodes using struct node* head = NULL
Declare variable n as nth to 3
Call function insert(10) to insert a node
Call display() to display the list
Call last(n) to find nth node from last of a list
Stop
Example Live demonstration#include<stdio.h> #include<stdlib.h> //structure of a node struct node{ int data; struct node *next; }*head,*temp; int count=0; //function for inserting nodes into a list void insert(int val){ struct node* newnode = (struct node*)malloc(sizeof(struct node)); newnode->data = val; newnode->next = NULL; if(head == NULL){ head = newnode; temp = head; count++; } else { temp->next=newnode; temp=temp->next; count++; } } //function for displaying a list void display(){ if(head==NULL) printf("no node "); else { temp=head; while(temp!=NULL) { printf("%d ",temp->data); temp=temp->next; } } } //function for finding 3rd node from the last of a linked list void last(int n){ int i; temp=head; for(i=0;i<count-n;i++){ temp=temp->next; } printf("</p><p>%drd node from the end of linked list is : %d" ,n,temp->data); } int main(){ //creating list struct node* head = NULL; int n=3; //inserting elements into a list insert(1); insert(2); insert(3); insert(4); insert(5); insert(6); //displaying the list printf("</p><p>linked list is : "); display(); //calling function for finding nth element in a list from last last(n); return 0; }Output
linked list is : 1 2 3 4 5 6
3rd node from the end of linked list is : 4
The above is the detailed content of In the C program, translate the following content into Chinese: Program to find the nth node from the bottom of a linked list. For more information, please follow other related articles on the PHP Chinese website!

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