Shift Last N elements to start
AppendLastNToFirst
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AppendLastNToFirst
The first line contains an Integer 't' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first line of each test case or query contains the elements of the singly linked list separated by a single space.
The second line contains the integer value 'N'. It denotes the number of nodes to be moved from last to the front of the singly linked list.
While specifying the list elements for input, -1 indicates the end of the singly linked list and hence, would never be a list element.
For each test case/query, print the resulting singly linked list of integers in a row, separated by a single space.
Output for every test case will be printed in a seperate line.
1 <= t <= 10^2
0 <= M <= 10^5
0 <= N < M
Time Limit: 1sec
Where 'M' is the size of the singly linked list.
2
1 2 3 4 5 -1
3
10 20 30 40 50 60 -1
5
3 4 5 1 2
20 30 40 50 60 10
1
10 6 77 90 61 67 100 -1
4
90 61 67 100 10 6 77
We have been required to move the last 4 nodes to the front of the list. Here, "90->61->67->100" is the list which represents the last 4 nodes. When we move this list to the front then the remaining part of the initial list which is, "10->6->77" is attached after 100. Hence, the new list formed with an updated head pointing to 90.
/****************************************************************
Following is the class structure of the Node class:
class Node
{
public:
int data;
Node *next;
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
*****************************************************************/
Node *appendLastNToFirst(Node *head, int n)
{
//Write your code here
if (n==0){
return head;
}
if (head==NULL){
return head;
}
int count=0;
int len=0;
Node* temp1=head;
Node* temp2=head;
while(temp1->next!=NULL){
temp1=temp1->next;
len++;
}
temp1->next=temp2;
while(count<len-n){
temp2=temp2->next;
count++;
}
head=temp2->next;
temp2->next=NULL;
return head;
}
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