leeecode 445. Add Two Numbers II 链表相加

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

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

题意很简单，做法如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>using namespace std;/*struct ListNode{    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};*/class Solution{public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)    {        vector<int> nums1, nums2;        while (l1)         {            nums1.push_back(l1->val);            l1 = l1->next;        }        while (l2)         {            nums2.push_back(l2->val);            l2 = l2->next;        }        int m = nums1.size(), n = nums2.size();        int sum = 0, carry = 0;        ListNode *head = nullptr, *p = nullptr;        for (int i = m - 1, j = n - 1; i >= 0 || j >= 0 || carry > 0; i--, j--)        {            sum = carry;            if (i >= 0)                sum += nums1[i];            if (j >= 0)                sum += nums2[j];            carry = sum / 10;            p = new ListNode(sum % 10);            p->next = head;            head = p;        }        return head;    }};