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.
Python
# Python program to subtract smaller # valued list from larger valued list # and return result as a list. # A linked List Node class Node: def __init__(self, new_data): self.data = new_data self.next = None # A utility which creates Node. def newNode(data): temp = Node(0) temp.data = data temp.next = None return temp # A utility function to get # length of linked list def getLength(Node): size = 0 while (Node != None): Node = Node.next size = size + 1 return size # A Utility that padds zeros in # front of the Node, with the # given diff def paddZeros(sNode, diff): if (sNode == None): return None zHead = newNode(0) diff = diff - 1 temp = zHead while (diff > 0): diff = diff - 1 temp.next = newNode(0) temp = temp.next temp.next = sNode return zHead borrow = True # 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. def subtractLinkedListHelper(l1, l2): global borrow if (l1 == None and l2 == None and not borrow ): return None l3 = None l4 = None if(l1 != None): l3 = l1.next if(l2 != None): l4 = l2.next previous = subtractLinkedListHelper(l3, l4) d1 = l1.data d2 = l2.data sub = 0 # If you have given the value value # to next digit then reduce the d1 by 1 if (borrow): d1 = d1 - 1 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 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. def subtractLinkedList(l1, l2): # Base Case. if (l1 == None and l2 == None): return None # In either of the case, get the # lengths of both # Linked list. len1 = getLength(l1) len2 = getLength(l2) lNode = None sNode = None temp1 = l1 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): if(len1 > len2): lNode = l1 else: lNode = l2 if(len1 > len2): sNode = l2 else: sNode = 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 != None and l2 != None): if (l1.data != l2.data): if(l1.data > l2.data ): lNode = temp1 else: lNode = temp2 if(l1.data > l2.data ): sNode = temp2 else: sNode = temp1 break l1 = l1.next l2 = l2.next global borrow # After calculating larger and smaller # Node, call subtractLinkedListHelper # which returns the subtracted # linked list. borrow = False return subtractLinkedListHelper(lNode, sNode) # A utility function to print # linked list def printList(Node): while (Node != None): print (Node.data, end =" ") Node = Node.next print(" ") # Driver code head1 = newNode(1) head1.next = newNode(0) head1.next.next = newNode(0) head2 = newNode(1) result = subtractLinkedList(head1, head2) printList(result) # This code is contributed by Arnab Kundu |
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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