2. Add Two Numbers

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本文介绍了一种使用链表表示非负整数并实现两数相加的方法。输入为两个逆序存储的链表，每个节点包含一个数字。通过遍历链表节点，逐位相加并考虑进位情况，最终返回表示和的新链表。

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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.

/**
 * 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) {
    ListNode * head, *p1, *p2 ,*p ,  *q;
        p1 = l1;
        p2 = l2;
        int sum , JingWei = 0 ;  // 保存每位之和  
        int a, b = 0 ; 
        head =q= NULL; 
        while (p1 != NULL || p2 != NULL)
        {
            if (p1 != NULL) a = p1->val;
            else  a = 0; 

            if (p2 != NULL) b = p2->val;
            else  b = 0; 

            sum = a + b + JingWei;
            if (sum >= 10)
            {
                sum -= 10; 
                JingWei = 1;
            }
            else JingWei = 0; 

            p = new ListNode(sum); 
            if (head == NULL)
            {
                head = p; 
            }
            else
            {
                q->next = p; 
            }
            q = p;
            if (p1!=NULL)   p1 = p1->next;
        if (p2 != NULL)   p2 = p2->next;
        }
        //最后还有进位
        if (JingWei)
        {
            p = new ListNode(JingWei);
            q->next = p;
        }
        return head;

    }
};