Print the last k nodes of the linked list in reverse order | Iterative Approaches

Given a linked list containing N nodes and a positive integer K where K should be less than or equal to N. The task is to print the last K nodes of the list in reverse order.

Examples:

Input : list: 1->2->3->4->5, K = 2
Output : 5 4

Input : list: 3->10->6->9->12->2->8, K = 4
Output : 8 2 12 9


The solution discussed in previous post uses recursive approach. The following article discusses two iterative approaches to solve the above problem.

Approach 1: The idea is to use stack data structure. Push all the linked list nodes data value to stack and pop first K elements and print them.

Below is the implementation of above approach:

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// C++ implementation to print the last k nodes
// of linked list in reverse order
#include <bits/stdc++.h>
using namespace std;
  
// Structure of a node
struct Node {
    int data;
    Node* next;
};
  
// Function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* newNode = new Node;
  
    // put in data
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}
  
// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head, int k)
{
    // if list is empty
    if (!head)
        return;
  
    // Stack to store data value of nodes.
    stack<int> st;
  
    // Push data value of nodes to stack
    while (head) {
        st.push(head->data);
        head = head->next;
    }
  
    int cnt = 0;
  
    // Pop first k elements of stack and
    // print them.
    while (cnt < k) {
        cout << st.top() << " ";
        st.pop();
        cnt++;
    }
}
  
// Driver code
int main()
{
    // Create list: 1->2->3->4->5
    Node* head = getNode(1);
    head->next = getNode(2);
    head->next->next = getNode(3);
    head->next->next->next = getNode(4);
    head->next->next->next->next = getNode(5);
  
    int k = 4;
  
    // print the last k nodes
    printLastKRev(head, k);
  
    return 0;
}

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Output:

5 4 3 2

Time Complexity: O(N)
Auxiliary Space: O(N)

The auxiliary space of the above approach can be reduced to O(k). The idea is to use two pointers. Place first pointer to beginning of the list and move second pointer to k-th node form beginning. Then find k-th node from end using approach discussed in this article: Find kth node from end of linked list. After finding kth node from end push all the remaining nodes in the stack. Pop all elements one by one from stack and print them.

Below is the implementation of the above efficient approach:

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// C++ implementation to print the last k nodes
// of linked list in reverse order
  
#include <bits/stdc++.h>
using namespace std;
  
// Structure of a node
struct Node {
    int data;
    Node* next;
};
  
// Function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* newNode = new Node;
  
    // put in data
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}
  
// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head, int k)
{
    // if list is empty
    if (!head)
        return;
  
    // Stack to store data value of nodes.
    stack<int> st;
  
    // Declare two pointers.
    Node *first = head, *sec = head;
  
    int cnt = 0;
  
    // Move second pointer to kth node.
    while (cnt < k) {
        sec = sec->next;
        cnt++;
    }
  
    // Move first pointer to kth node from end
    while (sec) {
        first = first->next;
        sec = sec->next;
    }
  
    // Push last k nodes in stack
    while (first) {
        st.push(first->data);
        first = first->next;
    }
  
    // Last k nodes are reversed when pushed
    // in stack. Pop all k elements of stack
    // and print them.
    while (!st.empty()) {
        cout << st.top() << " ";
        st.pop();
    }
}
  
// Driver code
int main()
{
    // Create list: 1->2->3->4->5
    Node* head = getNode(1);
    head->next = getNode(2);
    head->next->next = getNode(3);
    head->next->next->next = getNode(4);
    head->next->next->next->next = getNode(5);
  
    int k = 4;
  
    // print the last k nodes
    printLastKRev(head, k);
  
    return 0;
}

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Output:

5 4 3 2

Time Complexity: O(N)
Auxiliary Space: O(k)

Approach 2: The idea is to first reverse the linked list iteratively as discussed in following post: Reverse a linked list. After reversing print first k nodes of the reversed list. After printing restore the list by reversing the list again.

Below is the implementation of above approach:

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// C++ implementation to print the last k nodes
// of linked list in reverse order
#include <bits/stdc++.h>
using namespace std;
  
// Structure of a node
struct Node {
    int data;
    Node* next;
};
  
// Function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* newNode = new Node;
  
    // put in data
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}
  
// Function to reverse the linked list.
Node* reverseLL(Node* head)
{
    if (!head || !head->next)
        return head;
  
    Node *prev = NULL, *next = NULL, *curr = head;
  
    while (curr) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
  
    return prev;
}
  
// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head, int k)
{
    // if list is empty
    if (!head)
        return;
  
    // Reverse linked list.
    head = reverseLL(head);
  
    Node* curr = head;
  
    int cnt = 0;
  
    // Print first k nodes of linked list.
    while (cnt < k) {
        cout << curr->data << " ";
        cnt++;
        curr = curr->next;
    }
  
    // Restore the list.
    head = reverseLL(head);
}
  
// Driver code
int main()
{
    // Create list: 1->2->3->4->5
    Node* head = getNode(1);
    head->next = getNode(2);
    head->next->next = getNode(3);
    head->next->next->next = getNode(4);
    head->next->next->next->next = getNode(5);
  
    int k = 4;
  
    // print the last k nodes
    printLastKRev(head, k);
  
    return 0;
}

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Output:

5 4 3 2

Time Complexity: O(N)
Auxiliary Space: O(1)



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