n個節點的二叉樹n+1

Given a linked list and an integer n, append the last n elements of the LL to front. Assume given n will be smaller than length of LL.

給定一個鏈表和一個整數n，將LL的最后n個元素附加到前面。 假設給定的n將小于LL的長度。

Input format: Line 1: Linked list elements (separated by space and terminated by -1

輸入格式：第1行：鏈接的列表元素(以空格分隔并以-1終止

Sample Input 1 :1 2 3 4 5 -13Sample Output 1 :3 4 5 1 2

Description:

描述：

The question asks us to append the last N nodes to front, i.e the new linked list should first start from those N nodes and then traverse the rest of the nodes through the head of the old linked list.

這個問題要求我們將最后的N個節點附加到前面，即新的鏈表應首先從這N個節點開始，然后再通過舊鏈表的頭遍歷其余節點。

Example:

例：

For Linked List 1->2->3->4->5->6->NULLTo append the last 2 nodes, the new linked list should be:5->6->1->2->3->4->NULL

Solution Explanation:

解決方案說明：

To solve this problem, we take two pointers temp and t and point both of them to the head of the linked list. We take another variable i and equate it to – n. This i is used for finding out the head of the new linked list. Then we traverse the loop while temp != NULL. In the loop we check that if(i>=0) i.e temp is now n nodes away from t, t = t-> next. We will update i++ and temp = temp->next on each traversal. At last, we update temp-> next = head, head = t -> next and t-> next = NULL.

為了解決這個問題，我們使用兩個指針temp和t并將它們都指向鏈接列表的開頭。 我們采用另一個變量i并將其等于– n 。 我用于查找新鏈表的標題。 然后，我們在temp！= NULL時遍歷循環。 在循環中，我們檢查if(i> = 0)，即temp現在距離t距離n個節點， t = t-> next 。 我們將在每次遍歷時更新i ++和temp = temp-> next 。 最后，我們更新temp-> next = head ， head = t-> next和t-> next = NULL 。

Algorithm:

算法：

• STEP 1: Declare the function appendNNode with parameters (Node* head, int n)

  步驟1：使用參數聲明函數appendNNode (Node * head，int n)

• STEP 2: Declare two variables Node * temp , t and point both of them to head.

  步驟2：聲明兩個變量Node * temp ， t并將它們都指向head。

• STEP 3: Declare int i = -n

  步驟3：聲明int i = -n

• STEP 4: Repeat Step 5 and 6, while(temp->next != NULL)

  步驟4：重復步驟5和6， 同時(temp-> next！= NULL)

• STEP 5: if(i>=0) t = t-> next.

  步驟5： if(i> = 0)t = t-> next 。

• STEP 6: temp = temp-> next, i++.

  步驟6： temp = temp->接下來，i ++ 。

• STEP 7: temp->next = head, head = t->next, and t-> next =NULL

  步驟7： temp-> next = head ， head = t-> next和t-> next = NULL

• STEP 8: return head

  步驟8：返回頭

Steps:

腳步：

At first: 1->2->3->4->5->6->NULL, t->1 and temp->1.After complete traversal: 1->2->3->4->5->6->NULL, t->4 and temp->6.So, temp->next = head and head = t->nexti.e 5->6->1->2->3->4 --- (reconnecting to 5)Atlast, t-> next = NULLi.e 5->6->1->2->3->4->NULL .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

Function:

功能：

Node *appendNNodes(Node* head, int n){// Two pointers, one for traversal and // other for finding the new head of LLNode *temp = head, *t = head; //index maintained for finding new headint i = -n;while(temp->next!=NULL){//When temp went forward n nodes from tif(i>=0){ t = t->next;}temp = temp ->next;i++;}//Connecting the tail to headtemp->next = head; //Assigning the new nodehead = t->next; //Deleting the previous connectiont->next = NULL; return head; }

C++ Code:

C ++代碼：

#include<bits/stdc++.h>using namespace std;struct Node{// linked list Nodeint data;Node * next; };Node *newNode(int k){ //defining new nodeNode *temp = (Node*)malloc(sizeof(Node)); temp->data = k; temp->next = NULL; return temp; }//Used to add new node at the end of the list Node *addNode(Node* head, int k){if(head == NULL){head = newNode(k);}else{Node * temp = head;Node * node = newNode(k);while(temp->next!= NULL){temp = temp->next;}temp-> next = node;}return head; }// Used to create new linked list and return head Node *createNewLL(){int cont = 1;int data;Node* head = NULL;while(cont){cout<<"Enter the data of the Node"<<endl;cin>>data;head = addNode(head,data);cout<<"Do you want to continue?(0/1)"<<endl;cin>>cont;}return head; }//To print the Linked List void *printLL(Node * head){while(head!= NULL){cout<<head->data<<"->";head = head-> next;}cout<<"NULL"<<endl; }//Function Node *appendNNodes(Node* head, int n){// Two pointers, one for traversal and // other for finding the new head of LLNode *temp = head, *t = head; //index maintained for finding new headint i = -n; while(temp->next!=NULL){//When temp went forward n nodes from tif(i>=0){ t = t->next;}temp = temp ->next;i++;}//Connecting the tail to headtemp->next = head; //Assigning the new nodehead = t->next; //Deleting the previous connectiont->next = NULL; return head; }//Driver Main int main(){Node * head = createNewLL();cout<<"The linked list is"<<endl;printLL(head);int data;cout<<"Enter the number of nodes you want to append."<<endl;cin>>data;head = appendNNodes(head,data);cout<<"The new Linked List is" <<endl;printLL(head);return 0; } .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

Output

輸出量

Enter the data of the Node 1 Do you want to continue?(0/1) 1 Enter the data of the Node 2 Do you want to continue?(0/1) 1 Enter the data of the Node 3 Do you want to continue?(0/1) 1 Enter the data of the Node 4 Do you want to continue?(0/1) 1 Enter the data of the Node 5 Do you want to continue?(0/1) 1 Enter the data of the Node 6 Do you want to continue?(0/1) 1 Enter the data of the Node 7 Do you want to continue?(0/1) 0 The linked list is 1->2->3->4->5->6->7->NULL Enter the number of nodes you want to append. 3 The new Linked List is 5->6->7->1->2->3->4->NULL

翻譯自: https://www.includehelp.com/cpp-programs/append-last-n-nodes-to-first-in-the-linked-list.aspx

n個節點的二叉樹n+1

總結

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