You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
题意：链表形式的加法
思路：比较像大数加法吧，照着从左到右模拟过去
/**
* 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) {
if (l2 == NULL) return l1;
if (l1 == NULL) return l2;
ListNode *l3 = NULL;
ListNode *p = NULL;
int flag = 0;
while (l1 != NULL || l2 != NULL) {
int sum = flag;
if (l1 != NULL) {
sum += l1->val;
l1 = l1->next;
}
if (l2 != NULL) {
sum += l2->val;
l2 = l2->next;
}
int cur = sum % 10;
flag = sum / 10;
ListNode *tmp = new ListNode(cur);
if (l3 == NULL) {
l3 = tmp;
p = l3;
} else {
p->next = tmp;
p = p->next;
}
}
if (flag != 0) {
ListNode *tmp = new ListNode(flag);
p->next = tmp;
}
return l3;
}
};
原文：http://blog.csdn.net/u011345136/article/details/42320247