思路
一次遍历,短的后面当0补上即可,
时间复杂度$O(max(m,n))$
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *head = nullptr;
ListNode *tail = nullptr;
int carry = 0;
while(l1 || l2){
int n1 = l1 ? l1->val : 0;
int n2 = l2 ? l2->val : 0;
int sum = (n1 + n2 + carry);
carry = sum / 10 ;
if(head == nullptr) {
head = tail = new ListNode(sum % 10);
}
else {
tail->next = new ListNode(sum % 10);
tail = tail->next;
}
if(l1) l1 = l1->next;
if(l2) l2 = l2->next;
}
if(carry)tail->next = new ListNode(1);
return head;
}
};