LeetCode习题笔记——Add Two Numbers

这次问题看起来是两个数之和，其实主要考察两数相加中进位和对链表的一些基本操作，原题如下：

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

大意两个数字都是用链表反向保存（如2 -> 4 -> 3表示342），之后相加结果也用这个方式输出

解决方式其实并不困难，甚至题目这个反向保存让这个题目更容易解决，因为个位十位百位是反向的，所以我们遍历链表的时候就只用顺序遍历（相加的时候是从个位数开始相加，再判断进位，在进行下一位）。代码如下：

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        ListNode *head = (ListNode *)malloc(sizeof(ListNode));        ListNode *pre = head;        ListNode *node = NULL;        //进位        int c = 0,sum,val1,val2;        //加法        while(l1 != NULL || l2 != NULL || c != 0){            val1 = (l1 == NULL ? 0 : l1->val);            val2 = (l2 == NULL ? 0 : l2->val);            sum = val1 + val2 + c;            c = sum / 10;            node = (ListNode *)malloc(sizeof(ListNode));            node->val = sum % 10;            node->next = NULL;            //尾插法            pre->next = node;            pre = node;            l1 = (l1 == NULL ? NULL : l1->next);            l2 = (l2 == NULL ? NULL : l2->next);        }        return head->next;    }};