【LeetCode】2. Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

【思路】

利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

/**
* 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) {
if(l1==NULL && l2 == NULL) return NULL;
ListNode* head = new ListNode(0);
ListNode* tmp = head;
int add = 0;
while(l1 !=NULL || l2 !=NULL)
{
int val1 = 0;
if(l1!=NULL)
{
val1 = l1->val;
l1= l1->next;
}

int val2 = 0;
if(l2!=NULL)
{
val2 = l2->val;
l2= l2->next;
}

int sum = val1 + val2 + add;
tmp->next = new ListNode(sum % 10);
if(sum>=10)
add=1;
else add = 0;
tmp =tmp->next;

}
if(add==1)
tmp->next = new ListNode(1);
return head->next;
}
};