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;
}
};