Subtract Two Numbers represented as Linked Lists - GeeksforGeeks
Given two linked lists that represent two large positive numbers. Subtract the smaller number from 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 larger one.
It may be assumed that there are no extra leading zeros in input lists.
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Given two linked lists that represent two large positive numbers. Subtract the smaller number from 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 larger one.
It may be assumed that there are no extra leading zeros in input lists.
Input : l1 = 1 -> 0 -> 0 -> NULL, l2 = 1 -> NULL Output : 0->9->9->NULL
1) Calculate sizes of given two linked lists.
2) If sizes not are same, then append zeros in smaller linked list.
3) If size are same, then follow below steps:
….a) Find the smaller valued linked list.
….b) One by one subtract nodes of smaller sized linked list from larger size. Keep track of borrow while subtracting.
2) If sizes not are same, then append zeros in smaller linked list.
3) If size are same, then follow below steps:
….a) Find the smaller valued linked list.
….b) One by one subtract nodes of smaller sized linked list from larger size. Keep track of borrow while subtracting.
/* 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 = newNode(0); diff--; Node* temp = zHead; while (diff--) { temp->next = newNode(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, bool& borrow){ if (l1 == NULL && l2 == NULL && borrow == 0) return NULL; Node* previous = subtractLinkedListHelper(l1 ? l1->next : NULL, l2 ? l2->next : NULL, borrow); int d1 = l1->data; int d2 = l2->data; int 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 */ Node* current = newNode(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, 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 && l2) { if (l1->data != l2->data) { lNode = l1->data > l2->data ? temp1 : temp2; sNode = l1->data > l2->data ? temp2 : temp1; } l1 = l1->next; l2 = l2->next; } } // After calculating larger and smaller Node, call // subtractLinkedListHelper which returns the subtracted // linked list. bool borrow = false; return subtractLinkedListHelper(lNode, sNode, borrow);}