Add Two Numbers
Problem
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
解题思路
就是按照我们小时候的竖式加法的方式来计算就可以了。不同的是，这里的进位是向右进位；而我们小时候计算加法的进位是向左进位。
My Solution
class Solution
{
public:
ListNode* addTwoNumbers(ListNode *l1, ListNode *l2)
{
ListNode* p = l1;
ListNode* q = l2;
int sum = 0;
ListNode* sentinel = new ListNode(0);
ListNode* d = sentinel;
if ((p == NULL) && (q != NULL))
{
return q;
}
if ((p != NULL) && (q == NULL))
{
return p;
}
do
{
if (p != NULL)
{
sum += (p->val);
p = p->next;
}
else
{
sum += 0;
// p = p->next;
}
if (q != NULL)
{
sum += (q->val);
q = q->next;
}
else
{
sum += 0;
// q = q->next;
}
d->next = new ListNode((sum % 10));
d = d->next;
sum = (sum/10);
if(q==NULL && q==NULL && sum!=0)
{
d->next=new ListNode(sum);
}
}while (p != NULL || q != NULL);
return sentinel->next;
}
};本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
如有侵权请联系 cloudcommunity@tencent.com 删除
评论
登录后参与评论
推荐阅读
目录