# 0002. Add Two Numbers
###### tags: `Leetcode` `Medium` `Linked List` `Add Simulation`
Link: https://leetcode.com/problems/add-two-numbers/
## 思路
挺简单的一题，直接上code
方法一是新建了一个linked list存结果
方法二是直接用l2的空间存
## Code
### 方法一

/**
 * 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;
        if(!l1){
            return l2;
        }
        if(!l2){
            return l1;
        }
        while(l1||l2){
            int n1 = l1?l1->val:0;
            int n2 = l2?l2->val:0;
            int sum = n1+n2+carry;
            if(!head){
                head = tail = new ListNode(sum%10);
            }
            else{
                tail->next = new ListNode(sum%10);
                tail = tail->next;
            }
            carry = sum/10;
            if(l1){
                l1 = l1->next;
            }
            if(l2){
                l2 = l2->next;
            }
        }
        if(carry>0){
            tail->next = new ListNode(carry);
            tail = tail->next;
        }
        return head;
    }
};

### 方法二

/**
 * 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) {
        int carry = 0;
        if(!l1){
            return l2;
        }
        if(!l2){
            return l1;
        }
        ListNode *head = l2;
        ListNode *tail = nullptr;
        while(l1||l2){
            int n1 = l1?l1->val:0;
            int n2 = l2?l2->val:0;
            int sum = n1+n2+carry;
            if(l2){
                l2->val = sum%10;
                tail = l2;
            }
            else{
                if(!tail){
                    tail = new ListNode(sum%10);
                }
                else{
                    tail->next = new ListNode(sum%10);
                    // l1 = before->next;
                    tail = tail->next;
                }
            }
            
            carry = sum/10;
            if(l1){
                l1 = l1->next;
            }
            if(l2){
                l2 = l2->next;
            }
        }
        if(carry>0){
            tail->next = new ListNode(carry);
            tail = tail->next;
        }
        return head;
    }
};                                                                                                                                                                           

## Result
### 方法一
Runtime: 20 ms, faster than **91.37%** of C++ online submissions for Add Two Numbers.
Memory Usage: 71.3 MB, less than **17.68%** of C++ online submissions for Add Two Numbers.
### 方法二
Runtime: 28 ms, faster than **61.98%** of C++ online submissions for Add Two Numbers.
Memory Usage: 71 MB, less than **91.19%** of C++ online submissions for Add Two Numbers.