LeetCode: Add Two Numbers
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
模拟两个非负整数相加。注意进位，以及对应位之和为0时的情况即可。
AC代码：1
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84/**
* 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) {
bool l1ended = false;
bool l2ended = false;
int i1 = 0;
int i2 = 0;
int t = 0;
bool carry = false;
ListNode head(0);
ListNode* ptail = &head;
while(true)
{
if (l1)
{
i1 = l1->val;
l1 = l1->next;
}
else
{
i1 = 0;
l1ended = true;
}
if (l2)
{
i2 = l2->val;
l2 = l2->next;
}
else
{
i2 = 0;
l2ended = true;
}
if (carry)
{
t = i1 + i2 + 1;
}
else
{
t = i1 + i2;
}
if (t != 0)
{
if (t > 9) // 进位
{
t %= 10;
carry = true;
}
else
{
carry = false;
}
ListNode* n = new ListNode(t);
ptail->next = n;
ptail = n;
}
else
{
if (l1ended && l2ended)
{
break;
}
else
{
ListNode* n = new ListNode(0); // 有可能为0
ptail->next = n;
ptail = n;
}
}
}
return head.next;
}
};