【27】2. Add Two Numbers

2. Add Two Numbers

• Total Accepted: 241470
• Total Submissions: 910015
• Difficulty: Medium
• Contributors: Admin 

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.

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

/**
 * 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* dummy = new ListNode(-1);
        ListNode* tmp = dummy;
        int carry = 0;//carry要写在最外面 因为算value要加上去 但是第一步要算为0
        
        //while(l1 && l2){//分开写的话 l1，l2只有一个节点5的情况下 只会输出一个0 而不是0->1
        
        while(l1 || l2){//此时需要判断l1和l2 谁真谁假
            
            int n1 = l1 ? l1 -> val : 0;
            int n2 = l2 ? l2 -> val : 0;
        
            //int value = l1 -> val + l2 -> val + carry;
            int value = n1 + n2 + carry;
            carry = value / 10;//因为每个节点的都在10以内所以只可能和是两位数 所以不用while循环（value >= 10）
            
            ListNode* cur = new ListNode(value % 10);
            tmp -> next = cur;
            tmp = tmp -> next;
            
            //l1 = l1 -> next;
            //l2 = l2 -> next;//因为while循环的条件用的||，所以此时也需要判断
            
            if(l1){
                l1 = l1 -> next;
            }
            if(l2){
                l2 = l2 -> next;
            }
            
        }
        /*//用||把一条链长一条链短的情况包含了
        if(l1){
            tmp -> next = l1;
        }
        if(l2){
            tmp -> next = l2;
        }*/
        
        if(carry){
            tmp -> next = new ListNode(carry);
        }
        
        return dummy -> next;
    }
};