You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
Original Quiz in C++
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { } };
Solution 1 - C++
twonumbers.cpp
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ #define MNEXT(A) (A?(A->next?A->next:NULL):NULL) #define MSELECT(A,B) (A?A:(B?B:NULL)) class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if (!l1 && !l2) return NULL; ListNode *n=new ListNode(0); if (l1) n->val+=l1->val; if (l2) n->val+=l2->val; if (n->val>9) { ListNode *next; next=MNEXT(l1); if (!next) next=MNEXT(l2); if (next) // next node(s) exists, so carry over 1 to next digit { next->val += 1; } else // no next node found, so create new node with initial value=1 { ListNode *p=MSELECT(l1,l2); p->next=new ListNode(1); } n->val -=10; } if (MNEXT(l1) || MNEXT(l2)) n->next = addTwoNumbers(MNEXT(l1), MNEXT(l2)); return n; } };
Solution 2 in Java
add_two_numbers.java
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode p = l1, q = l2, curr = dummyHead; int carry = 0; while (p != null || q != null) { int x = (p != null) ? p.val : 0; int y = (q != null) ? q.val : 0; int sum = carry + x + y; carry = sum / 10; curr.next = new ListNode(sum % 10); curr = curr.next; if (p != null) p = p.next; if (q != null) q = q.next; } if (carry > 0) { curr.next = new ListNode(carry); } return dummyHead.next; }