Add two numbers represented by linked lists | Set 2 - GeeksforGeeks
// Adds two linked lists of same size represented by head1 and head2 and returns
http://yuanhsh.iteye.com/blog/2181045
Given two numbers represented by two linked lists, write a function that returns sum list. The sum list is linked list representation of addition of two input numbers. It is not allowed to modify the lists. Also, not allowed to use explicit extra space (Hint: Use Recursion).
Example
Input: First List: 5->6->3 // represents number 563 Second List: 8->4->2 // represents number 842 Output Resultant list: 1->4->0->5 // represents number 1405
1) Calculate sizes of given two linked lists.
2) If sizes are same, then calculate sum using recursion. Hold all nodes in recursion call stack till the rightmost node, calculate sum of rightmost nodes and forward carry to left side.
3) If size is not same, then follow below steps:
….a) Calculate difference of sizes of two linked lists. Let the difference be diff
….b) Move diff nodes ahead in the bigger linked list. Now use step 2 to calculate sum of smaller list and right sub-list (of same size) of larger list. Also, store the carry of this sum.
….c) Calculate sum of the carry (calculated in previous step) with the remaining left sub-list of larger list. Nodes of this sum are added at the beginning of sum list obtained previous step.
2) If sizes are same, then calculate sum using recursion. Hold all nodes in recursion call stack till the rightmost node, calculate sum of rightmost nodes and forward carry to left side.
3) If size is not same, then follow below steps:
….a) Calculate difference of sizes of two linked lists. Let the difference be diff
….b) Move diff nodes ahead in the bigger linked list. Now use step 2 to calculate sum of smaller list and right sub-list (of same size) of larger list. Also, store the carry of this sum.
….c) Calculate sum of the carry (calculated in previous step) with the remaining left sub-list of larger list. Nodes of this sum are added at the beginning of sum list obtained previous step.
// Adds two linked lists of same size represented by head1 and head2 and returns
// head of the resultant linked list. Carry is propagated while returning from// the recursionnode* addSameSize(node* head1, node* head2, int* carry){ // Since the function assumes linked lists are of same size, // check any of the two head pointers if (head1 == NULL) return NULL; int sum; // Allocate memory for sum node of current two nodes node* result = (node *)malloc(sizeof(node)); // Recursively add remaining nodes and get the carry result->next = addSameSize(head1->next, head2->next, carry); // add digits of current nodes and propagated carry sum = head1->data + head2->data + *carry; *carry = sum / 10; sum = sum % 10; // Assigne the sum to current node of resultant list result->data = sum; return result;}// This function is called after the smaller list is added to the bigger// lists's sublist of same size. Once the right sublist is added, the carry// must be added toe left side of larger list to get the final result.void addCarryToRemaining(node* head1, node* cur, int* carry, node** result){ int sum; // If diff. number of nodes are not traversed, add carry if (head1 != cur) { addCarryToRemaining(head1->next, cur, carry, result); sum = head1->data + *carry; *carry = sum/10; sum %= 10; // add this node to the front of the result push(result, sum); }}// The main function that adds two linked lists represented by head1 and head2.// The sum of two lists is stored in a list referred by resultvoid addList(node* head1, node* head2, node** result){ node *cur; // first list is empty if (head1 == NULL) { *result = head2; return; } // second list is empty else if (head2 == NULL) { *result = head1; return; } int size1 = getSize(head1); int size2 = getSize(head2) ; int carry = 0; // Add same size lists if (size1 == size2) *result = addSameSize(head1, head2, &carry); else { int diff = abs(size1 - size2); // First list should always be larger than second list. // If not, swap pointers if (size1 < size2) swapPointer(&head1, &head2); // move diff. number of nodes in first list for (cur = head1; diff--; cur = cur->next); // get addition of same size lists *result = addSameSize(cur, head2, &carry); // get addition of remaining first list and carry addCarryToRemaining(head1, cur, &carry, result); } // if some carry is still there, add a new node to the fron of // the result list. e.g. 999 and 87 if (carry) push(result, carry);}http://yuanhsh.iteye.com/blog/2181045
- private int carry = 0;
- public ListNode addReversedLinkedList(ListNode a, ListNode b) {
- if(a==null) {
- return b;
- } else if(b==null) {
- return a;
- }
- int m = this.getListSize(a);
- int n = this.getListSize(b);
- ListNode result;
- if(m==n) {
- result = this.addListWithSameSize(a, b);
- } else {
- int diff = Math.abs(m-n);
- ListNode list1 = m>n?a:b; // make sure the first list is larger
- ListNode list2 = m>n?b:a;
- ListNode cur = list1;
- while(diff-->0) cur = cur.next;
- result = this.addListWithSameSize(cur, list2);
- result = this.addRemainingList(list1, cur, result);
- }
- if(this.carry>0) {
- ListNode newRes = new ListNode(0);
- newRes.val = this.carry;
- newRes.next = result;
- return newRes;
- }
- return result;
- }
- private int getListSize(ListNode n) {
- int size = 0;
- while(n!= null) {
- n = n.next;
- size++;
- }
- return size;
- }
- private ListNode addListWithSameSize(ListNode a, ListNode b) {
- if(a==null || b==null) return null;
- ListNode result = new ListNode(0);
- result.next = addListWithSameSize(a.next, b.next);
- int sum = a.val+b.val+this.carry;
- this.carry = sum/10;
- result.val = sum%10;
- return result;
- }
- private ListNode addRemainingList(ListNode list1, ListNode cur, ListNode result) {
- if(list1==cur) return result;
- ListNode next = addRemainingList(list1.next, cur, result);
- int sum = list1.val + this.carry;
- this.carry = sum/10;
- ListNode ret = new ListNode(0);
- ret.val = sum%10;
- ret.next = next;
- return ret;
- }