n个节点的二叉树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.

给定一个链表和一个整数n，将LL的最后n个元素附加到前面。 假设给定的n将小于LL的长度。

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

输入格式：第1行：链接的列表元素(以空格分隔并以-1终止

    Sample Input 1 :
1 2 3 4 5 -1
3
Sample 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.

这个问题要求我们将最后的N个节点附加到前面，即新的链表应首先从这N个节点开始，然后再通过旧链表的头遍历其余节点。

Example:

例：

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

Solution Explanation:

解决方案说明：

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.

为了解决这个问题，我们使用两个指针temp和t并将它们都指向链接列表的开头。 我们采用另一个变量i并将其等于– n 。 我用于查找新链表的标题。 然后，我们在temp！= NULL时遍历循环。 在循环中，我们检查if(i> = 0)，即temp现在距离t距离n个节点， t = t-> next 。 我们将在每次遍历时更新i ++和temp = temp-> next 。 最后，我们更新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)

  步骤1：使用参数声明函数appendNNode (Node * head，int n)

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

  步骤2：声明两个变量Node * temp ， t并将它们都指向head。

• STEP 3: Declare int i = -n

  步骤3：声明int i = -n

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

  步骤4：重复步骤5和6， 同时(temp-> next！= NULL)

• STEP 5: if(i>=0) t = t-> next.

  步骤5： if(i> = 0)t = t-> next 。

• STEP 6: temp = temp-> next, i++.

  步骤6： temp = temp->接下来，i ++ 。

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

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

• STEP 8: return head

  步骤8：返回头

Steps:

脚步：

    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->next
i.e 5->6->1->2->3->4 --- (reconnecting to 5)
Atlast, t-> next = NULL
i.e 5->6->1->2->3->4->NULL

Function:

功能：

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

C++ Code:

C ++代码：

#include<bits/stdc++.h>
using namespace std;
struct Node{// linked list Node
int data;
Node * next;
};
Node *newNode(int k){ //defining new node
Node *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 LL
Node *temp = head, *t = head;
//index maintained for finding new head
int i = -n;
while(temp->next!=NULL){//When temp went forward n nodes from t
if(i>=0){
t = t->next;
}
temp = temp ->next;
i++;
}
//Connecting the tail to head
temp->next = head;
//Assigning the new node
head = t->next;
//Deleting the previous connection
t->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;
}

Output

输出量

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

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

n个节点的二叉树n+1