LeetCode 2. 两数相加（单链表反转）

题目链接：https://leetcode-cn.com/problems/add-two-numbers/

给出两个非空的链表用来表示两个非负的整数。其中，它们各自的位数是按照逆序的方式存储的，并且它们的每个节点只能存储一位数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

考查单链表反转

/*** 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) {int sum, one = 0;ListNode *nextNode = NULL;while(l1 != NULL || l2 != NULL){if(l1 != NULL && l2 != NULL){sum = l1->val + l2->val + one;one = sum/10;ListNode *newNode = new ListNode(sum%10);newNode->next = nextNode;nextNode = newNode;l1 = l1->next;l2 = l2->next;}else if(l1 != NULL && l2 == NULL){sum = l1->val + one;one = sum/10;ListNode *newNode = new ListNode(sum%10);newNode->next = nextNode;nextNode = newNode;l1 = l1->next;}else{sum = l2->val + one;one = sum/10;ListNode *newNode = new ListNode(sum%10);newNode->next = nextNode;nextNode = newNode;l2 = l2->next;}}if(one != 0){ListNode *newNode = new ListNode(1);newNode->next = nextNode;nextNode = newNode;}return reverseList(nextNode);}ListNode* reverseList(ListNode *cur){if(cur->next == NULL)return cur;ListNode *prevNode = NULL, *nextNode = cur->next;while(cur != NULL && cur->next != NULL){cur->next = prevNode;prevNode = cur;cur = nextNode;nextNode = nextNode->next;}cur->next = prevNode;return cur;//反转链表后的新的头结点}
};

直接对每位进行相加

class Solution {	// 2020.10.4
public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {int carry = 0, sum;ListNode* newhead = NULL, *prev = NULL;while(l1 || l2 || carry){sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;ListNode* node = new ListNode(sum%10);carry = sum/10;if(l1)l1 = l1->next;if(l2)l2 = l2->next;if(!newhead){newhead = node;prev = node;}else{prev->next = node;prev = node;}}return newhead;}
};