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#2 Add Two Numbers

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Description

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:

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

解法 进位链表法

从两个链表头结点开始依次相加，记录进位，供下一次相加使用，直到其中一个链表结束，将另一个链表的内容加进位输出到结果高位即可，对链表的要求高

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/**
 * 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 *p=l1,*q=l2; // 工作指针
        int carry = 0; // 进位
        ListNode *l3=new ListNode(-1); // 头结点
        ListNode *r=l3; // l3的工作指针
        int temp;

        while (p||q){
            if(p&&q){
                temp = p->val + q->val + carry;
                carry = temp/10;
                ListNode *node=new ListNode(temp%10);
                r->next = node;
                r = node;
                p = p->next;
                q = q->next;
            }
        
            if(p&&!q){
                temp = p->val + carry;
                carry = temp/10;
                ListNode *node=new ListNode(temp%10);
                r->next = node;
                r = node;
                p = p->next;
            }

            if(!p&&q){
                temp = q->val + carry;
                carry = temp/10;
                ListNode *node=new ListNode(temp%10);
                r->next = node;
                r = node;
                q = q->next;
            }
        }

        if(carry>0){
            ListNode *node=new ListNode(carry);
            r->next = node;
            r = node;
        }
        return l3->next;
    }
};