给定 n 个节点,任务是打印链表末尾的第 n 个节点。程序不得更改列表中节点的顺序,而应仅打印链表最后一个节点中的第 n 个节点。
Input -: 10 20 30 40 50 60 N=3 Output -: 40
在上面的例子中,从第一个节点开始,遍历到 count-n 个节点,即 10,20 30,40, 50,60,所以倒数第三个节点是 40。
而不是如此高效地遍历整个列表可以遵循的方法 -
如果我们使用这种方法,计数将为 5,程序将迭代循环直到 5-3,即 2,因此从 0th 位置上的 10 开始,直到 20结果是在第 1 个位置,第 30 个位置在第二个位置。因此,通过这种方法,不需要遍历整个列表直到结束,这将节省空间和内存。
Start 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
现场演示
#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; }
linked list is : 1 2 3 4 5 6 3rd node from the end of linked list is : 4
以上是在C程序中,将以下内容翻译为中文:查找链表倒数第n个节点的程序的详细内容。更多信息请关注PHP中文网其他相关文章!