两个整数分别用两个Linked List 表示 ，把它们加起来

2. Add Two Numbers.cpp

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int before = 0;
        ListNode* res = new ListNode(0);
        ListNode* pre = res;
        while (l1 != NULL || l2 != NULL){
            int current = 0;
            if (!l2) { current = l1->val; l1 = l1->next; }
            else if (!l1) { current = l2->val;  l2 = l2->next; }
            else {
                current = l1->val + l2->val;
                l1 = l1->next;
                l2 = l2->next;
            }
            current += before;
            if (current > 9){
                current -= 10;
                before = 1;
            }
            else before = 0;
            pre->next = new ListNode(current);
            pre = pre->next;
        }
        if (before != 0) pre->next = new ListNode(before);
        return res->next;
            
    }
};

More concise solution.cpp

/**
 * 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) {
        int before = 0;
        ListNode* res = new ListNode(0);
        ListNode* pre = res;
        while (l1 || l2 || before){
            if (l1) { before += l1->val; l1 = l1->next; }
            if (l2) { before += l2->val;  l2 = l2->next; }
            pre->next = new ListNode(before%10);
            before /= 10;
            pre = pre->next;
        }
        return res->next;
            
    }
};

可以把进位放到while的判断中去。
求进位的时候可以用pre/10 pre%10来求。