leetcode-【中等题】2. Add Two Numbers

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