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# Java Program To Find The Sum Of Last N Nodes Of The Given Linked List

Given a linked list and a number n. Find the sum of the last n nodes of the linked list.
Constraints: 0 <= n <= number of nodes in the linked list.

Examples:

```Input: 10->6->8->4->12, n = 2
Output: 16
Sum of last two nodes:
12 + 4 = 16

Input: 15->7->9->5->16->14, n = 4
Output: 44```

Method 1: (Recursive approach using system call stack)
Recursively traverse the linked list up to the end. Now during the return from the function calls, add up the last n nodes. The sum can be accumulated in some variable passed by reference to the function or to some global variable.

## Java

 `// Java implementation to find the sum of` `// last 'n' nodes of the Linked List` `import` `java.util.*;`   `class` `GFG{` `    `  `// A Linked list node ` `static` `class` `Node` `{` `    ``int` `data;` `    ``Node next;` `};` `static` `Node head; ` `static` `int` `n, sum;`   `// Function to insert a node at the` `// beginning of the linked list` `static` `void` `push(Node head_ref, ` `                 ``int` `new_data)` `{` `    ``// Allocate node ` `    ``Node new_node = ``new` `Node();` `    `  `    ``// Put in the data ` `    ``new_node.data = new_data;` `    `  `    ``// Link the old list to the ` `    ``// new node ` `    ``new_node.next = head_ref;` `    `  `    ``// Move the head to point to the ` `    ``// new node ` `    ``head_ref = new_node;` `    ``head = head_ref;` `}`   `// Function to recursively find the sum of last` `// 'n' nodes of the given linked list` `static` `void` `sumOfLastN_Nodes(Node head)` `{` `    ``// if head = NULL` `    ``if` `(head == ``null``)` `        ``return``;`   `    ``// Recursively traverse the remaining ` `    ``// nodes` `    ``sumOfLastN_Nodes(head.next);`   `    ``// if node count 'n' is greater than 0` `    ``if` `(n > ``0``) ` `    ``{` `        ``// Accumulate sum` `        ``sum = sum + head.data;`   `        ``// Reduce node count 'n' by 1` `        ``--n;` `    ``}` `}`   `// Utility function to find the sum of ` `// last 'n' nodes` `static` `int` `sumOfLastN_NodesUtil(Node head, ` `                                ``int` `n)` `{` `    ``// if n == 0` `    ``if` `(n <= ``0``)` `        ``return` `0``;`   `    ``sum = ``0``;`   `    ``// Find the sum of last 'n' nodes` `    ``sumOfLastN_Nodes(head);`   `    ``// Required sum` `    ``return` `sum;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args) ` `{` `    ``head = ``null``;`   `    ``// Create linked list 10.6.8.4.12` `    ``push(head, ``12``);` `    ``push(head, ``4``);` `    ``push(head, ``8``);` `    ``push(head, ``6``);` `    ``push(head, ``10``);`   `    ``n = ``2``;` `    ``System.out.print(``"Sum of last "` `+ n + ` `                     ``" nodes = "` `+ ` `                     ``sumOfLastN_NodesUtil(head, n));` `}` `}` `// This code is contributed by 29AjayKumar`

Output:

`Sum of last 2 nodes = 16`

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(n), if system call stack is being considered.

Method 2: (Iterative approach using user-defined stack)
It is an iterative procedure to the recursive approach explained in Method 1 of this post. Traverses the nodes from left to right. While traversing pushes the nodes to a user-defined stack. Then pops the top n values from the stack and adds them.

## Java

 `// Java implementation to find the sum of last` `// 'n' nodes of the Linked List` `import` `java.util.*;`   `class` `GFG{`   `// A Linked list node ` `static` `class` `Node ` `{` `    ``int` `data;` `    ``Node next;` `};`   `// Function to insert a node at the` `// beginning of the linked list` `static` `Node push(Node head_ref, ` `                 ``int` `new_data)` `{` `    ``// Allocate node ` `    ``Node new_node = ``new` `Node();` `    `  `    ``// Put in the data ` `    ``new_node.data = new_data;` `    `  `    ``// Link the old list to the new node ` `    ``new_node.next = head_ref;` `    `  `    ``// Move the head to point to the ` `    ``// new node ` `    ``head_ref = new_node;` `    ``return` `head_ref;` `}`   `// Utility function to find the sum of ` `// last 'n' nodes` `static` `int` `sumOfLastN_NodesUtil(Node head, ` `                                ``int` `n)` `{` `    ``// if n == 0` `    ``if` `(n <= ``0``)` `        ``return` `0``;`   `    ``Stack st = ``new` `Stack();` `    ``int` `sum = ``0``;`   `    ``// Traverses the list from left to right` `    ``while` `(head != ``null``) ` `    ``{` `        ``// Push the node's data onto the ` `        ``// stack 'st'` `        ``st.push(head.data);`   `        ``// Move to next node` `        ``head = head.next;` `    ``}`   `    ``// Pop 'n' nodes from 'st' and` `    ``// add them` `    ``while` `(n-- >``0``)` `    ``{` `        ``sum += st.peek();` `        ``st.pop();` `    ``}`   `    ``// Required sum` `    ``return` `sum;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``Node head = ``null``;`   `    ``// Create linked list 10.6.8.4.12` `    ``head = push(head, ``12``);` `    ``head = push(head, ``4``);` `    ``head = push(head, ``8``);` `    ``head = push(head, ``6``);` `    ``head = push(head, ``10``);`   `    ``int` `n = ``2``;` `    ``System.out.print(``"Sum of last "` `+ n + ` `                     ``" nodes = "` `+ ` `                     ``sumOfLastN_NodesUtil(head, n));` `}` `}` `// This code is contributed by 29AjayKumar`

Output:

`Sum of last 2 nodes = 16`

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(n), stack size

Method 3: (Reversing the linked list)
Following are the steps:

1. Reverse the given linked list.
2. Traverse the first n nodes of the reversed linked list.
4. Reverse the linked list back to its original order.

## Java

 `// Java implementation to find the sum of last` `// 'n' nodes of the Linked List` `import` `java.util.*;`   `class` `GFG{` `    `  `// A Linked list node ` `static` `class` `Node` `{` `    ``int` `data;` `    ``Node next;` `};` `static` `Node head; `   `// Function to insert a node at the` `// beginning of the linked list` `static` `void` `push(Node head_ref, ` `                 ``int` `new_data)` `{` `    ``// Allocate node ` `    ``Node new_node = ``new` `Node();` `    `  `    ``// Put in the data ` `    ``new_node.data = new_data;` `    `  `    ``// Link the old list to the new node ` `    ``new_node.next = head_ref;` `    `  `    ``// Move the head to point to the ` `    ``// new node ` `    ``head_ref = new_node;` `    ``head=head_ref;` `}`   `static` `void` `reverseList(Node head_ref)` `{` `    ``Node current, prev, next;` `    ``current = head_ref;` `    ``prev = ``null``;`   `    ``while` `(current != ``null``) ` `    ``{` `        ``next = current.next;` `        ``current.next = prev;` `        ``prev = current;` `        ``current = next;` `    ``}`   `    ``head_ref = prev;` `    ``head = head_ref;` `}`   `// Utility function to find the sum of ` `// last 'n' nodes` `static` `int` `sumOfLastN_NodesUtil(``int` `n)` `{` `    ``// if n == 0` `    ``if` `(n <= ``0``)` `        ``return` `0``;`   `    ``// Reverse the linked list` `    ``reverseList(head);`   `    ``int` `sum = ``0``;` `    ``Node current = head;`   `    ``// Traverse the 1st 'n' nodes of the ` `    ``// reversed linked list and add them` `    ``while` `(current != ``null` `&& n-- >``0``) ` `    ``{  ` `        ``// Accumulate node's data to 'sum'` `        ``sum += current.data;`   `        ``// Move to next node` `        ``current = current.next;` `    ``}`   `    ``// Reverse back the linked list` `    ``reverseList(head);`   `    ``// Required sum` `    ``return` `sum;` `}`   `// Driver code` `public` `static` `void` `main(String[] args) ` `{`   `    ``// Create linked list 10.6.8.4.12` `    ``push(head, ``12``);` `    ``push(head, ``4``);` `    ``push(head, ``8``);` `    ``push(head, ``6``);` `    ``push(head, ``10``);`   `    ``int` `n = ``2``;` `    ``System.out.println(``"Sum of last "` `+ n + ` `                       ``" nodes = "` `+ ` `                       ``sumOfLastN_NodesUtil(n));` `}` `}` `// This code is contributed by PrinciRaj1992 `

Output:

`Sum of last 2 nodes = 16`

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)

Method 4: (Using the length of linked list)
Following are the steps:

1. Calculate the length of the given Linked List. Let it be len.
2. First, traverse the (len – n) nodes from the beginning.
3. Then traverse the remaining n nodes and while traversing add them.

## Java

 `// Java implementation to find the sum of last ` `// 'n' nodes of the Linked List ` `class` `GFG{`   `// A Linked list node ` `static` `class` `Node` `{ ` `    ``int` `data; ` `    ``Node next; ` `}; ` `static` `Node head;`   `// Function to insert a node at the ` `// beginning of the linked list ` `static` `void` `push(Node head_ref, ` `                 ``int` `new_data) ` `{ ` `    ``// Allocate node ` `    ``Node new_node = ``new` `Node(); `   `    ``// Put in the data ` `    ``new_node.data = new_data; `   `    ``// Link the old list to the new node ` `    ``new_node.next = head_ref; `   `    ``// Move the head to point to the ` `    ``// new node ` `    ``head_ref = new_node; ` `    ``head = head_ref;` `} `   `// Utility function to find the sum of` `// last 'n' nodes ` `static` `int` `sumOfLastN_NodesUtil(Node head, ` `                                ``int` `n) ` `{ ` `    ``// if n == 0 ` `    ``if` `(n <= ``0``) ` `        ``return` `0``; `   `    ``int` `sum = ``0``, len = ``0``; ` `    ``Node temp = head; `   `    ``// Calculate the length of the ` `    ``// linked list ` `    ``while` `(temp != ``null``)` `    ``{ ` `        ``len++; ` `        ``temp = temp.next; ` `    ``} `   `    ``// Count of first (len - n) nodes ` `    ``int` `c = len - n; ` `    ``temp = head; `   `    ``// Just traverse the 1st 'c' nodes ` `    ``while` `(temp != ``null``&&c-- >``0``)` `    ``{                     ` `        ``// Move to next node ` `        ``temp = temp.next; ` `    ``}` `    `  `    ``// Now traverse the last 'n' nodes and ` `    ``// add them ` `    ``while` `(temp != ``null``) ` `    ``{ ` `        ``// Accumulate node's data to sum ` `        ``sum += temp.data; `   `        ``// Move to next node ` `        ``temp = temp.next; ` `    ``} `   `    ``// Required sum ` `    ``return` `sum; ` `} `   `// Driver code ` `public` `static` `void` `main(String[] args)` `{` `    ``// Create linked list 10.6.8.4.12 ` `    ``push(head, ``12``); ` `    ``push(head, ``4``); ` `    ``push(head, ``8``); ` `    ``push(head, ``6``); ` `    ``push(head, ``10``); `   `    ``int` `n = ``2``; ` `    ``System.out.println(``"Sum of last "` `+ n + ` `                       ``" nodes = "` `+ ` `                       ``sumOfLastN_NodesUtil(head, n)); ` `}` `}` `// This code is contributed by 29AjayKumar`

Output:

`Sum of last 2 nodes = 16`

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)

Method 5: (Use of two pointers requires single traversal)
Maintain two pointers – reference pointer and main pointer. Initialize both reference and main pointers to head. First, move reference pointer to n nodes from head and while traversing accumulate node’s data to some variable, say sum. Now move both pointers simultaneously until the reference pointer reaches the end of the list and while traversing accumulate all node’s data to sum pointed by the reference pointer and accumulate all node’s data to some variable, say, temp, pointed by the main pointer. Now, (sum – temp) is the required sum of the last n nodes.

## Java

 `// Java implementation to find the sum of last` `// 'n' nodes of the Linked List` `class` `GfG` `{` `    ``// Defining structure` `    ``static` `class` `Node` `    ``{` `        ``int` `data;` `        ``Node next;` `    ``}`   `    ``static` `Node head;`   `    ``static` `void` `printList(Node start)` `    ``{` `        ``Node temp = start;` `        ``while` `(temp != ``null``)` `        ``{` `            ``System.out.print(temp.data + ``" "``);` `            ``temp = temp.next;` `        ``}` `        ``System.out.println();` `    ``}`   `    ``// Push function` `    ``static` `void` `push(Node start, ``int` `info)` `    ``{` `        ``// Allocating node` `        ``Node node = ``new` `Node();`   `        ``// Info into node` `        ``node.data = info;`   `        ``// Next of new node to head` `        ``node.next = start;`   `        ``// head points to new node` `        ``head = node;` `    ``}`   `    ``private` `static` `int` `sumOfLastN_NodesUtil(Node head, ` `                                            ``int` `n)` `    ``{` `        ``// if n == 0` `        ``if` `(n <= ``0``)` `            ``return` `0``;`   `        ``int` `sum = ``0``, temp = ``0``;` `        ``Node ref_ptr, main_ptr;` `        ``ref_ptr = main_ptr = head;`   `        ``// Traverse 1st 'n' nodes through 'ref_ptr' ` `        ``// and accumulate all node's data to 'sum'` `        ``while` `(ref_ptr != ``null` `&& (n--) > ``0``)` `        ``{` `            ``sum += ref_ptr.data;`   `            ``// Move to next node` `            ``ref_ptr = ref_ptr.next;` `        ``}`   `        ``// Traverse to the end of the linked list` `        ``while` `(ref_ptr != ``null``)` `        ``{` `            ``// Accumulate all node's data to 'temp' ` `            ``// pointed by the 'main_ptr'` `            ``temp += main_ptr.data;`   `            ``// Accumulate all node's data to 'sum' ` `            ``// pointed by the 'ref_ptr'` `            ``sum += ref_ptr.data;`   `            ``// Move both the pointers to their ` `            ``// respective next nodes` `            ``main_ptr = main_ptr.next;` `            ``ref_ptr = ref_ptr.next;` `        ``}`   `        ``// Required sum` `        ``return` `(sum - temp);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``head = ``null``;`   `        ``// Adding elements to Linked List` `        ``push(head, ``12``);` `        ``push(head, ``4``);` `        ``push(head, ``8``);` `        ``push(head, ``6``);` `        ``push(head, ``10``);` `        ``printList(head);` `        ``int` `n = ``2``;` `        ``System.out.println(``"Sum of last "` `+ n + ` `                           ``" nodes = "` `+ ` `                           ``sumOfLastN_NodesUtil(head, n));` `    ``}` `}` `// This code is contributed by shubham96301`

Output:

`Sum of last 2 nodes = 16`

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)

Please refer complete article on Find the sum of last n nodes of the given Linked List for more details!

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