题目描述
给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。
如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外，这两个数都不会以 0 开头。
示例
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807
解题思路
题解
遍历相加注意进位即可。
C++
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == nullptr) return l2;
if (l2 == nullptr) return l1;
ListNode *iter = new ListNode(-1);
ListNode *head = iter;
int carry = 0;
while (l1 != nullptr and l2 != nullptr)
{
carry = l1->val + l2->val + carry;
ListNode *node = new ListNode(carry % 10);
iter->next = node;
iter = node;
carry = carry / 10;
l1 = l1->next;
l2 = l2->next;
}
while (l1 != nullptr)
{
carry = l1->val + carry;
ListNode *node = new ListNode(carry % 10);
iter->next = node;
iter = node;
carry = carry / 10;
l1 = l1->next;
}
while (l2 != nullptr)
{
carry = l2->val + carry;
ListNode *node = new ListNode(carry % 10);
iter->next = node;
iter = node;
carry = carry / 10;
l2 = l2->next;
}
if (carry > 0)
{
ListNode *node = new ListNode(carry);
iter->next = node;
}
return head->next;
}
};
python
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
if not l1:
return l2
if not l2:
return l1
carry = 0
head = ListNode(-1)
iters = head
while(l1 and l2):
carry = l1.val + l2.val + carry
node = ListNode(carry % 10)
carry = carry // 10
iters.next = node
iters = node
l1 = l1.next
l2 = l2.next
while(l1):
carry = l1.val + carry
node = ListNode(carry % 10)
carry = carry // 10
iters.next = node
iters = node
l1 = l1.next
while(l2):
carry = l2.val + carry
node = ListNode(carry % 10)
carry = carry // 10
iters.next = node
iters = node
l2 = l2.next
if carry > 0:
node = ListNode(carry)
iters.next = node
return head.next
