2.Add two numbers

最新推荐文章于 2021-12-02 14:05:46 发布

原创最新推荐文章于 2021-12-02 14:05:46 发布 · 109 阅读

0 ·

CC 4.0 BY-SA版权

leetcode算法专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种算法，用于解决两个非空链表表示的非负整数相加的问题。链表中的数字以逆序存储，每个节点包含一个数字。通过示例解释了如何将(2->4->3)和(5->6->4)相加得到7->0->8的过程。

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.

/**
* 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) {
ListNode* head1 = l1;
ListNode* head2 = l2;
ListNode* head = NULL;
ListNode* prev = NULL;
int carry = 0;

while(head1 || head2)
{
int v1 = head1?head1->val:0;
int v2 = head2?head2->val:0;
int tmp = v1 + v2 + carry;

carry = tmp / 10;
tmp = tmp % 10;
ListNode * cur_node = new ListNode(tmp);
if(!head)
{
head = cur_node;
}
if(prev)
{
prev->next = cur_node;
}
prev = cur_node;

head1 = head1?head1->next:NULL;
head2 = head2?head2->next:NULL;
}
if(carry)
{
ListNode * cur_node = new ListNode(carry);
prev->next = cur_node;
}
return head;
}
};