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.
- Calculate sizes of given two linked lists.
- If sizes are not the same, then append zeros in the smaller linked list.
- If the size is the same, then follow the below steps:
- Find the smaller valued linked list.
- 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.
Javascript
<script> // Javascript program to subtract smaller // valued list from larger valued list // and return result as a list. // head of list var head; var borrow; // Node Class class Node { // Constructor to create a // new node constructor(d) { this.data = d; this.next = null; } } /* A utility function to get length of linked list */function getLength(node) { var 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 */function paddZeros(sNode, diff) { if (sNode == null) return null; var zHead = new Node(0); diff--; var 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. */function subtractLinkedListHelper(l1, l2) { if (l1 == null && l2 == null && borrow == false) return null; var previous = subtractLinkedListHelper((l1 != null) ? l1.next : null, (l2 != null) ? l2.next : null); var d1 = l1.data; var d2 = l2.data; var sub = 0; /* if you have given the value 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 var 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. */function subtractLinkedList(l1, l2) { // Base Case. if (l1 == null && l2 == null) return null; // In either of the case, get the // lengths of both Linked list. var len1 = getLength(l1); var len2 = getLength(l2); var lNode = null, sNode = null; var temp1 = l1; var 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 function printList(head) { var temp = head; while (temp != null) { document.write(temp.data + " "); temp = temp.next; } } // Driver code var head = new Node(1); head.next = new Node(0); head.next.next = new Node(0); var head2 = new Node(1); var result = subtractLinkedList(head, head2); printList(result); // This code contributed by aashish1995 </script> |
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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