leetcode-2.Add Two Numbers

来源：互联网发布：淘宝网太阳镜编辑：程序博客网时间：2024/06/02 02:42

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

class Solution {public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {ListNode* result_head = new ListNode(0);  // new指向的内存块不会随着函数结束释放掉int carry = 0;ListNode* curr = result_head;while (l1 != NULL || l2 != NULL){int x = (l1 != NULL) ? l1->val : 0;int y = (l2 != NULL) ? l2->val : 0;int sum = x + y + carry;carry = sum / 10;curr->next =new ListNode(sum % 10);curr = curr->next;if (l1 != NULL)l1 = l1->next;if (l2 != NULL)l2 = l2->next;}if (carry > 0)curr->next = new ListNode(carry);return result_head->next;}};

关于在局部函数中使用new及 malloc的知识，以下链接的回答非常清楚：

https://zhidao.baidu.com/question/392722638371416605.html

0 0