代码下载网站,河北恒山建设集团网站,网站百度云,网上写作文的网站n个节点的二叉树n1Given 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#xff0c;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 form…n个节点的二叉树n1Given 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(i0) 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 * headint 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(i0) 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
.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 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(i0){
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 #includebits/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){
coutEnter the data of the Nodeendl;
cindata;
head addNode(head,data);
coutDo you want to continue?(0/1)endl;
cincont;
}
return head;
}
//To print the Linked List
void *printLL(Node * head){
while(head! NULL){
couthead-data-;
head head- next;
}
coutNULLendl;
}
//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(i0){
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();
coutThe linked list isendl;
printLL(head);
int data;
coutEnter the number of nodes you want to append.endl;
cindata;
head appendNNodes(head,data);
coutThe 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 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.aspxn个节点的二叉树n1
