Given two numbers represented by two linked lists, write a function that returns the sum list. The sum list is linked list representation of the addition of two input numbers. It is not allowed to modify the lists. Also, not allowed to use explicit extra space (Hint: Use Recursion).
Example :
Input: First List: 5->6->3 Second List: 8->4->2 Output: Resultant list: 1->4->0->5
We have discussed a solution here which is for linked lists where a least significant digit is the first node of lists and the most significant digit is the last node. In this problem, the most significant node is the first node and the least significant digit is the last node and we are not allowed to modify the lists. Recursion is used here to calculate sum from right to left.
Following are the steps.
- Calculate sizes of given two linked lists.
- If sizes are same, then calculate sum using recursion. Hold all nodes in recursion call stack till the rightmost node, calculate the sum of rightmost nodes and forward carry to the left side.
- If size is not same, then follow below steps:
- Calculate difference of sizes of two linked lists. Let the difference be diff.
- Move diff nodes ahead in the bigger linked list. Now use step 2 to calculate the sum of the smaller list and right sub-list (of the same size) of a larger list. Also, store the carry of this sum.
- Calculate the sum of the carry (calculated in the previous step) with the remaining left sub-list of a larger list. Nodes of this sum are added at the beginning of the sum list obtained the previous step.
Below is a dry run of the above approach:
Below image is the implementation of the above approach.
Java
// A Java recursive program to add // two linked listspublic class linkedlistATN { class node { int val; node next; public node(int val) { this.val = val; } } // Function to print linked list void printlist(node head) { while (head != null) { System.out.print( head.val + " "); head = head.next; } } node head1, head2, result; int carry; /* A utility function to push a value to linked list */ void push(int val, int list) { node newnode = new node(val); if (list == 1) { newnode.next = head1; head1 = newnode; } else if (list == 2) { newnode.next = head2; head2 = newnode; } else { newnode.next = result; result = newnode; } } // Adds two linked lists of same size // represented by head1 and head2 and // returns head of the resultant linked // list. Carry is propagated while // returning from the recursion void addsamesize(node n, node m) { // Since the function assumes // linked lists are of same // size, check any of the two // head pointers if (n == null) return; // Recursively add remaining nodes // and get the carry addsamesize(n.next, m.next); // Add digits of current nodes and // propagated carry int sum = n.val + m.val + carry; carry = sum / 10; sum = sum % 10; // Push this to result list push(sum, 3); } node cur; // This function is called after the // smaller list is added to the bigger // lists's sublist of same size. Once // the right sublist is added, the carry // must be added to the left side of // larger list to get the final result. void propogatecarry(node head1) { // If diff. number of nodes are // not traversed, add carry if (head1 != cur) { propogatecarry(head1.next); int sum = carry + head1.val; carry = sum / 10; sum %= 10; // Add this node to the front // of the result push(sum, 3); } } int getsize(node head) { int count = 0; while (head != null) { count++; head = head.next; } return count; } // The main function that adds two // linked lists represented by head1 // and head2. The sum of two lists is // stored in a list referred by result void addlists() { // first list is empty if (head1 == null) { result = head2; return; } // first list is empty if (head2 == null) { result = head1; return; } int size1 = getsize(head1); int size2 = getsize(head2); // Add same size lists if (size1 == size2) { addsamesize(head1, head2); } else { // First list should always be // larger than second list. If // not, swap pointers if (size1 < size2) { node temp = head1; head1 = head2; head2 = temp; } int diff = Math.abs(size1 - size2); // Move diff. number of nodes in // first list node temp = head1; while (diff-- >= 0) { cur = temp; temp = temp.next; } // Get addition of same size lists addsamesize(cur, head2); // Get addition of remaining first // list and carry propogatecarry(head1); } // If some carry is still there, add // a new node to the front of the // result list. e.g. 999 and 87 if (carry > 0) push(carry, 3); } // Driver code public static void main(String args[]) { linkedlistATN list = new linkedlistATN(); list.head1 = null; list.head2 = null; list.result = null; list.carry = 0; int arr1[] = { 9, 9, 9 }; int arr2[] = { 1, 8 }; // Create first list as 9->9->9 for (int i = arr1.length - 1; i >= 0; --i) list.push(arr1[i], 1); // Create second list as 1->8 for (int i = arr2.length - 1; i >= 0; --i) list.push(arr2[i], 2); list.addlists(); list.printlist(list.result); }}// This code is contributed by Rishabh Mahrsee |
Output:
1 0 1 7
Time Complexity:
O(m+n) where m and n are the sizes of given two linked lists.
Auxiliary Space: O(n+m) due to recursive stack space
Iterative Approach: This implementation does not have any recursion call overhead, which means it is an iterative solution. Since we need to start adding numbers from the last of the two linked lists. So, here we will use the stack data structure to implement this.
- We will firstly make two stacks from the given two linked lists.
- Then, we will run a loop till both the stack become empty.
- in every iteration, we keep the track of the carry.
- In the end, if carry>0, that means we need an extra node at the start of the resultant list to accommodate this carry.
Java
// Java Iterative program to add // two linked lists import java.io.*;import java.util.*;class GFG{ static class Node{ int data; Node next; public Node(int data) { this.data = data; }}static Node l1, l2, result;// To push a new node to linked listpublic static void push(int new_data){ // Allocate node Node new_node = new Node(0); // Put in the data new_node.data = new_data; // Link the old list of the // new node new_node.next = l1; // Move the head to point to the // new node l1 = new_node;}public static void push1(int new_data){ // Allocate node Node new_node = new Node(0); // Put in the data new_node.data = new_data; // Link the old list of the // new node new_node.next = l2; // Move the head to point to // the new node l2 = new_node;}// To add two new numberspublic static Node addTwoNumbers(){ Stack<Integer> stack1 = new Stack<>(); Stack<Integer> stack2 = new Stack<>(); while (l1 != null) { stack1.add(l1.data); l1 = l1.next; } while (l2 != null) { stack2.add(l2.data); l2 = l2.next; } int carry = 0; Node result = null; while (!stack1.isEmpty() || !stack2.isEmpty()) { int a = 0, b = 0; if (!stack1.isEmpty()) { a = stack1.pop(); } if (!stack2.isEmpty()) { b = stack2.pop(); } int total = a + b + carry; Node temp = new Node(total % 10); carry = total / 10; if (result == null) { result = temp; } else { temp.next = result; result = temp; } } if (carry != 0) { Node temp = new Node(carry); temp.next = result; result = temp; } return result;}// To print a linked listpublic static void printList(){ while (result != null) { System.out.print(result.data + " "); result = result.next; } System.out.println();}// Driver codepublic static void main(String[] args){ int arr1[] = {5, 6, 7}; int arr2[] = {1, 8}; int size1 = 3; int size2 = 2; // Create first list as 5->6->7 int i; for(i = size1 - 1; i >= 0; --i) push(arr1[i]); // Create second list as 1->8 for(i = size2 - 1; i >= 0; --i) push1(arr2[i]); result = addTwoNumbers(); printList();}}// This code is contributed by RohitOberoi |
Output:
5 8 5
Time Complexity: O(m + n) where m and n are the number of nodes in the given two linked lists.
Auxiliary Space: O(m+n), where m and n are the number of nodes in the given two linked lists.
Related Article: Add two numbers represented by linked lists | Set 1 Please refer complete article on Add two numbers represented by linked lists | Set 2 for more details!
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