# Leetcode 2. Add Two Numbers (C語言)
- 題目
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.
- 範例
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    if(!l1)return l2;
    if(!l2)return l1;
    struct ListNode *root,*current,*Nnode;
    current=(struct ListNode*)malloc(sizeof(struct ListNode));
    root=current;
    int v1,v2,carry=0,temp;
    while(l1||l2){
        v1=l1?l1->val:0;
        v2=l2?l2->val:0;
        temp=v1+v2+carry;
        if(temp>9){
            temp%=10;
            carry=1;
        }
        else{
            carry=0;
        }
        Nnode=(struct ListNode*)malloc(sizeof(struct ListNode));
        Nnode->val=temp;
        current->next=Nnode;
        current=current->next;
        if(l1)l1=l1->next;
        if(l2)l2=l2->next;
    }
    if(carry==1){
        Nnode=(struct ListNode*)malloc(sizeof(struct ListNode));
        Nnode->val=1;
        Nnode->next=NULL;
        current->next=Nnode;
    }
    Nnode->next=NULL;
    current=root->next;
    free(root);
    return current;
}

思路:v1+v2+carry(進位)=該位置之值，每次加完判斷有無進位。
可能情況有三種
1. l1沒了l2還有，l2補完
2. l2沒了l1還有，l1補完
3. l1還有l2還有，相加。
若末端仍進位則額外宣告一點記憶體空間來存放。