LeetCode | 2) Add Two Numbers

题目

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

/**
* 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) {

}
};

思路

只使用O(1)的额外空间，将相加的和存到

l1

中，返回

l1

链表头。

代码

class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* head{l1};
int carry{0};
int value{0};
while(l1->next != nullptr && l2->next != nullptr)
{
value = l1->val + l2->val + carry;
carry = value / 10;
value %= 10;
l1->val = value;
l1 = l1->next;
l2 = l2->next;
}
//碰到l1或者l2的最后一个节点
value = l1->val + l2->val + carry;
carry = value / 10;
value %= 10;
l1->val = value;
//如果l2的后继非空，则修改l1->next指向l2的后继
if ((l2 = l2->next) != nullptr)
{
l1->next = l2;
}
while (carry && l1->next)
{
//当carry非0 并且l1的后继非空，则转移l1至它的后继
l1 = l1->next;
value = l1->val + carry;
carry = value / 10;
value %= 10;
l1->val = value;
}
if (carry)
{
l1->next = new ListNode(carry);
}
return head;
}
};

运行时间约30ms，不需要创建新节点，没有多余的代码，不知道为什么这么慢。。