Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

解题思路：设置一个进位变量carry来记录进位，剩下的主要是指针的操作，& 加一级指针，* 降一级指针。

class Solution {
  
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int x=0, y=0, carry=0, sum=0;
        ListNode *h=NULL, **t=&h;
        
        while (l1!=NULL || l2!=NULL){
            x = getValueAndMoveNext(l1);
            y = getValueAndMoveNext(l2);
            
            sum = carry + x + y;
            
            ListNode *node = new ListNode(sum%10);
            *t = node;//*t降一级，为指向ListNode的指针。
            t = (&node->next);//t是指向下一个ListNode的指针的指针。
            
            carry = sum/10;
        }
        
        if (carry > 0) {//最后还有进位的话，重新建立一个ListNode。
            ListNode *node = new ListNode(carry%10);
            *t = node;
        }
        
        return h;
    }
private:
    int getValueAndMoveNext(ListNode* &l){
        int x = 0;
        if (l != NULL){
            x = l->val;
            l = l->next;
        }
        return x;
    }
};

Add Two Numbers

原文：http://blog.csdn.net/ruzhuxiaogu/article/details/47381337