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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

普通解法

这道题重在思路，编程没有什么c++技巧。我的思路如下：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode*l3=new ListNode(0);
        ListNode *head=l3;
        int carry=0;
        while(l1!=NULL and l2!=NULL){

            if(l1->val+l2->val+carry<10){
                l3->val=l1->val+l2->val+carry;
                carry=0;
            }
            else {
                l3->val=(l1->val+l2->val+carry)%10;
                carry=(l1->val+l2->val+carry)/10;
            }

            l1=l1->next;
            l2=l2->next;
            if(carry!=0 or l1!=NULL or l2!=NULL){
                 l3->next=new ListNode(0);
                l3=l3->next;
                l3->val=carry;
            }

        }
        while(l1){
            if(l1->val+carry<10){
                l3->val=l1->val+carry;
                carry=0;
            }
            else {
                l3->val=(l1->val+carry)%10;
                carry=(l1->val+carry)/10;
            }
            l1=l1->next;
            if(carry!=0 or l1!=NULL) {
                l3->next = new ListNode(0);
                l3 = l3->next;
                l3->val=carry;
            }

            }
        while(l2){

            if(l2->val+carry<10){
                l3->val=l2->val+carry;
                carry=0;
            }
            else {
                l3->val=(l2->val+carry)%10;
                carry=(l2->val+carry)/10;
            }
            l2=l2->next;
            if(carry!=0 or l2!=NULL) {
                l3->next = new ListNode(0);
                l3 = l3->next;
                l3->val=carry;
            }

        }

        return head;

    }

我将本题，分为两种情况

1. 两个列表长度相同时：依次计算，然后单独考虑进位情况
2. 两个列表长度不同时：先用1方法计算，然后再用循环来遍历剩下的链表

while(l1){
            if(l1->val+carry<10){
                l3->val=l1->val+carry;
                carry=0;
            }
            else {
                l3->val=(l1->val+carry)%10;
                carry=(l1->val+carry)/10;
            }
            l1=l1->next;
            if(carry!=0 or l1!=NULL) {
                l3->next = new ListNode(0);
                l3 = l3->next;
                l3->val=carry;
            }

            }

高级解法

高级解法，时间复杂度上并没有提高，但代码很精简，利用了递归解决问题：

  public ListNode addTwoNumbers2(ListNode l1, ListNode l2) {
        if (l1 == null || l2 == null) {
            return l1 == null ? l2 : l1;
        }
        int value = l1.val + l2.val;
        ListNode result = new ListNode(value % 10);
        result.next = addTwoNumbers2(l1.next, l2.next);
        if (value >= 10) {
            result.next = addTwoNumbers2(new ListNode(value / 10), result.next);
        }
        return result;
    }

这里不推荐用递归方法求解，因为牺牲的代价太大了