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/**
* 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 hand 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;
}
};
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Solution 2 in Java
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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;
} |