Given two numbers represented by two lists, write a function that returns the sum list. The sum list is a list representation of the addition of two input numbers.
Example:
Input:
List1: 5->6->3 // represents number 563
List2: 8->4->2 // represents number 842
Output:
Resultant list: 1->4->0->5 // represents number 1405
Explanation: 563 + 842 = 1405 Input:
List1: 7->5->9->4->6 // represents number 75946
List2: 8->4 // represents number 84
Output:
Resultant list: 7->6->0->3->0// represents number 76030
Explanation: 75946+84=76030
Approach: Traverse both lists and One by one pick nodes of both lists and add the values. If the sum is more than 10 then make carry as 1 and reduce sum. If one list has more elements than the other then consider the remaining values of this list as 0.
The steps are:
- Traverse the two linked lists from start to end
- Add the two digits each from respective linked lists.
- If one of the lists has reached the end then take 0 as its digit.
- Continue it until both the end of the lists.
- If the sum of two digits is greater than 9 then set carry as 1 and the current digit as sum % 10
Below is the implementation of this approach.
<script> // Javascript program to add two numbers // represented by linked list var head1, head2;
class Node { constructor(val)
{
this .data = val;
this .next = null ;
}
} /* Adds contents of two linked lists and return the head node of resultant
list */
function addTwoLists(first, second)
{ // res is head node of the resultant
// list
var res = null ;
var prev = null ;
var temp = null ;
var carry = 0, sum;
// while both lists exist
while (first != null ||
second != null )
{
// Calculate value of next digit in
// resultant list. The next digit is
// sum of following things
// (i) Carry
// (ii) Next digit of first list (if
// there is a next digit)
// (ii) Next digit of second list (if
// there is a next digit)
sum = carry + (first != null ? first.data : 0) +
(second != null ? second.data : 0);
// Update carry for next calculation
carry = (sum >= 10) ? 1 : 0;
// Update sum if it is greater than 10
sum = sum % 10;
// Create a new node with sum as data
temp = new Node(sum);
// If this is the first node then set
// it as head of the resultant list
if (res == null )
{
res = temp;
}
// If this is not the first node then
// connect it to the rest.
else {
prev.next = temp;
}
// Set prev for next insertion
prev = temp;
// Move first and second pointers to
// next nodes
if (first != null )
{
first = first.next;
}
if (second != null )
{
second = second.next;
}
}
if (carry > 0)
{
temp.next = new Node(carry);
}
// return head of the resultant list
return res;
} /* Utility function to print a linked list */
function printList(head)
{ while (head != null )
{
document.write(head.data + " " );
head = head.next;
}
document.write( "<br/>" );
} // Driver Code // Creating first list head1 = new Node(7);
head1.next = new Node(5);
head1.next.next = new Node(9);
head1.next.next.next = new Node(4);
head1.next.next.next.next = new Node(6);
document.write( "First List is " );
printList(head1); // Creating second list head2 = new Node(8);
head2.next = new Node(4);
document.write( "Second List is " );
printList(head2); // Add the two lists and see the // result rs = addTwoLists(head1, head2); document.write( "Resultant List is " );
printList(rs); // This code is contributed by aashish1995 </script> |
Output:
First List is 7 5 9 4 6 Second List is 8 4 Resultant list is 5 0 0 5 6
Complexity Analysis:
-
Time Complexity: O(m + n), where m and n are numbers of nodes in first and second lists respectively.
The lists need to be traversed only once. -
Space Complexity: O(m + n).
A temporary linked list is needed to store the output number
Related Article: Add two numbers represented by linked lists | Set 2
Please refer complete article on Add two numbers represented by linked lists | Set 1 for more details!