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# Java Program To Subtract Two Numbers Represented As Linked Lists

Given two linked lists that represent two large positive numbers. Subtract the smaller number from the larger one and return the difference as a linked list. Note that the input lists may be in any order, but we always need to subtract smaller from the larger ones.
It may be assumed that there are no extra leading zeros in input lists.
Examples:

```Input: l1 = 1 -> 0 -> 0 -> NULL,  l2 = 1 -> NULL
Output: 0->9->9->NULL
Explanation: Number represented as
lists are 100 and 1, so 100 - 1 is 099

Input: l1 = 7-> 8 -> 6 -> NULL,  l2 = 7 -> 8 -> 9 NULL
Output: 3->NULL
Explanation: Number represented as
lists are 786 and  789, so 789 - 786 is 3,
as the smaller value is subtracted from
the larger one.```

Approach: Following are the steps.

1. Calculate sizes of given two linked lists.
2. If sizes are not the same, then append zeros in the smaller linked list.
3. If the size is the same, then follow the below steps:
1. Find the smaller valued linked list.
2. One by one subtract nodes of the smaller-sized linked list from the larger size. Keep track of borrow while subtracting.

Following is the implementation of the above approach.

## Java

 `// Java program to subtract smaller valued ` `// list from larger valued list and return ` `// result as a list. ` `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` `  `class` `LinkedList  ` `{ ` `    ``// Head of list ` `    ``static` `Node head;  ` `    ``boolean` `borrow; ` ` `  `    ``// Node Class  ` `    ``static` `class` `Node  ` `    ``{ ` `        ``int` `data; ` `        ``Node next; ` ` `  `        ``// Constructor to create a  ` `        ``// new node ` `        ``Node(``int` `d) ` `        ``{ ` `            ``data = d; ` `            ``next = ``null``; ` `        ``} ` `    ``} ` ` `  `    ``/* A utility function to get length  ` `       ``of linked list */` `    ``int` `getLength(Node node) ` `    ``{ ` `        ``int` `size = ``0``; ` `        ``while` `(node != ``null``)  ` `        ``{ ` `            ``node = node.next; ` `            ``size++; ` `        ``} ` `        ``return` `size; ` `    ``} ` ` `  `    ``/* A Utility that padds zeros in front  ` `       ``of the Node, with the given diff */` `    ``Node paddZeros(Node sNode, ``int` `diff) ` `    ``{ ` `        ``if` `(sNode == ``null``) ` `            ``return` `null``; ` ` `  `        ``Node zHead = ``new` `Node(``0``); ` `        ``diff--; ` `        ``Node temp = zHead; ` `        ``while` `((diff--) != ``0``)  ` `        ``{ ` `            ``temp.next = ``new` `Node(``0``); ` `            ``temp = temp.next; ` `        ``} ` `        ``temp.next = sNode; ` `        ``return` `zHead; ` `    ``} ` ` `  `    ``/* Subtract LinkedList Helper is a recursive ` `       ``function, move till the last Node, and  ` `       ``subtract the digits and create the Node  ` `       ``and return the Node. If d1 < d2, we borrow  ` `       ``the number from previous digit. */` `    ``Node subtractLinkedListHelper(Node l1,  ` `                                  ``Node l2) ` `    ``{ ` `        ``if` `(l1 == ``null` `&&  ` `            ``l2 == ``null` `&& borrow == ``false``) ` `            ``return` `null``; ` ` `  `        ``Node previous = subtractLinkedListHelper( ` `                        ``(l1 != ``null``) ? l1.next: ``null``, ` `                        ``(l2 != ``null``) ? l2.next : ``null``); ` ` `  `        ``int` `d1 = l1.data; ` `        ``int` `d2 = l2.data; ` `        ``int` `sub = ``0``; ` ` `  `        ``/* If you have given the value to  ` `           ``next digit then reduce the d1 by 1 */` `        ``if` `(borrow)  ` `        ``{ ` `            ``d1--; ` `            ``borrow = ``false``; ` `        ``} ` ` `  `        ``/* If d1 < d2, then borrow the number  ` `           ``from previous digit. Add 10 to d1  ` `           ``and set borrow = true; */` `        ``if` `(d1 < d2)  ` `        ``{ ` `            ``borrow = ``true``; ` `            ``d1 = d1 + ``10``; ` `        ``} ` ` `  `        ``// Subtract the digits  ` `        ``sub = d1 - d2; ` ` `  `        ``// Create a Node with sub value  ` `        ``Node current = ``new` `Node(sub); ` ` `  `        ``// Set the Next pointer as Previous  ` `        ``current.next = previous; ` ` `  `        ``return` `current; ` `    ``} ` ` `  `    ``/* This API subtracts two linked lists  ` `       ``and returns the linked list which  ` `       ``shall have the subtracted result. */` `    ``Node subtractLinkedList(Node l1,  ` `                            ``Node l2) ` `    ``{ ` `        ``// Base Case. ` `        ``if` `(l1 == ``null` `&& l2 == ``null``) ` `            ``return` `null``; ` ` `  `        ``// In either of the case, get the  ` `        ``// lengths of both Linked list. ` `        ``int` `len1 = getLength(l1); ` `        ``int` `len2 = getLength(l2); ` ` `  `        ``Node lNode = ``null``, sNode = ``null``; ` ` `  `        ``Node temp1 = l1; ` `        ``Node temp2 = l2; ` ` `  `        ``// If lengths differ, calculate the  ` `        ``// smaller Node and padd zeros for  ` `        ``// smaller Node and ensure both larger  ` `        ``// Node and smaller Node has equal length. ` `        ``if` `(len1 != len2)  ` `        ``{ ` `            ``lNode = len1 > len2 ? l1 : l2; ` `            ``sNode = len1 > len2 ? l2 : l1; ` `            ``sNode = paddZeros(sNode, Math.abs(len1 - len2)); ` `        ``} ` ` `  `        ``else` `{ ` `            ``// If both list lengths are equal, then ` `            ``// calculate the larger and smaller list. ` `            ``// If 5-6-7 & 5-6-8 are linked list, then ` `            ``// walk through linked list at last Node ` `            ``// as 7 < 8, larger Node is 5-6-8 and ` `            ``// smaller Node is 5-6-7. ` `            ``while` `(l1 != ``null` `&& l2 != ``null``)  ` `            ``{ ` `                ``if` `(l1.data != l2.data)  ` `                ``{ ` `                    ``lNode = (l1.data > l2.data ?  ` `                             ``temp1 : temp2); ` `                    ``sNode = (l1.data > l2.data ?  ` `                             ``temp2 : temp1); ` `                    ``break``; ` `                ``} ` `                ``l1 = l1.next; ` `                ``l2 = l2.next; ` `            ``} ` `        ``} ` ` `  `        ``// After calculating larger and smaller  ` `        ``// Node, call subtractLinkedListHelper  ` `        ``// which returns the subtracted linked list. ` `        ``borrow = ``false``; ` `        ``return` `subtractLinkedListHelper(lNode,  ` `                                        ``sNode); ` `    ``} ` ` `  `    ``// Function to display the linked list ` `    ``static` `void` `printList(Node head) ` `    ``{ ` `        ``Node temp = head; ` `        ``while` `(temp != ``null``)  ` `        ``{ ` `            ``System.out.print(temp.data +  ` `                             ``" "``); ` `            ``temp = temp.next; ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``Node head = ``new` `Node(``1``); ` `        ``head.next = ``new` `Node(``0``); ` `        ``head.next.next = ``new` `Node(``0``); ` `        ``Node head2 = ``new` `Node(``1``); ` `        ``LinkedList ob = ``new` `LinkedList(); ` `        ``Node result = ob.subtractLinkedList(head,  ` `                                            ``head2); ` `        ``printList(result); ` `    ``} ` `} ` `// This article is contributed by Chhavi `

Output:

```0 9 9
```

Complexity Analysis:

• Time complexity: O(n).
As no nested traversal of linked list is needed.
• Auxiliary Space: O(n).
If recursive stack space is taken into consideration O(n) space is needed.

Please refer complete article on Subtract Two Numbers represented as Linked Lists for more details!

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