leetcode-【中等题】2. Add Two Numbers

题目

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

链接

https://leetcode.com/problems/add-two-numbers/

答案

1、直接按顺序加就行，保存进位carry

2、到最后还需要考虑进位carry是否为0

代码

/**
 * 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) {
        if(l1 == NULL)
        {
            return l2;
        }
        
        if(l2 == NULL)
        {
            return l1;
        }
        
        ListNode *ans = NULL;
        ListNode *point = NULL;
        int carry = 0;
        int sum = 0;
        while(l1 != NULL && l2 != NULL)
        {
            sum = carry + l1->val + l2->val;
            carry = sum / 10;
            sum = sum % 10;
            
            ListNode *value = new ListNode(sum);
            if(ans == NULL)
            {
                ans = value;
            }
            
            if(point != NULL)
            {
                point->next = value;
            }
            point = value;
            
            l1 = l1->next;
            l2 = l2->next;
        }
        
        while(l1 != NULL)
        {
            sum = carry + l1->val;
            carry = sum / 10;
            sum = sum % 10;
            ListNode *value = new ListNode(sum);
            point->next = value;
            point = value;
            
            l1 = l1->next;
        }
        
        while(l2 != NULL)
        {
            sum = carry + l2->val;
            carry = sum / 10;
            sum = sum % 10;
            ListNode *value = new ListNode(sum);
            point->next = value;
            point = value;
            
            l2 = l2->next;
        }
        
        if(carry != 0)
        {
            ListNode *value = new ListNode(carry);
            point->next = value;
            point = value;
        }
        
        return ans;
    }
};