leetcode-algorithms-2 Add Two Numbers
leetcode-algorithms-2 Add Two Numbers
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.
解法
同时遍历两个链表,对值进行相加.得到的值%10就是新的链表的值,同时存下/10(只可能是1)的值,用于下个位数的增值.
/**
* 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 *l = nullptr;
ListNode *lt = nullptr;
ListNode *node1 = l1;
ListNode *node2 = l2;
int lastaddval = 0;
while(true)
{
if (node1 == nullptr && node2 == nullptr) break;
int nodeval1 = 0;
int nodeval2 = 0;
if (node1 != nullptr)
{
nodeval1 = node1->val;
node1 = node1->next;
}
if (node2 != nullptr)
{
nodeval2 = node2->val;
node2 = node2->next;
}
int sumval = nodeval1 + nodeval2 + lastaddval;
lastaddval = sumval / 10;
if (l == nullptr)
{
l = new ListNode(sumval % 10);
lt = l;
}
else
{
ListNode *n = new ListNode(sumval % 10);
lt->next = n;
lt = n;
}
}
if (lastaddval != 0)
{
ListNode *n = new ListNode(lastaddval);
lt->next = n;
lt = n;
}
return l;
}
};
时间复杂度: O(max(m,n)).m和n分别是l1和l2的长度.
空间复杂度: O(max(m,n)).
