Add Two Numbers II问题及解法

问题描述：

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

示例:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

问题分析：

两数相加问题，注意考虑溢出问题，这里采用字符串的加减法。

过程详见代码：

/**
* 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) {
string res1, res2;
while (l1)
{
res1.push_back(l1->val + '0');
l1 = l1->next;
}
while (l2)
{
res2.push_back(l2->val + '0');
l2 = l2->next;
}
int flag = 0,t;
int i = res1.size() - 1,j = res2.size()- 1;
ListNode * curNode, *head = nullptr;
while (i >= 0 || j >= 0)
{
if (i >= 0 && j >= 0)
{
t = res1[i] - '0' + res2[j] - '0' + flag;
i--;
j--;
}
else if (i >= 0)
{
t = res1[i] - '0' + flag;
i--;
}
else
{
t = res2[j] - '0' + flag;
j--;
}
flag = t / 10;
curNode = new ListNode(t % 10);
curNode->next = head;
head = curNode;
}
if (flag)
{
curNode = new ListNode(flag);
curNode->next = head;
head = curNode;
}
return head;
}
};