002 Add Two Numbers [Leetcode]

题目内容：

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

今天开始进入链表专题。

题目不难，只要按照链表节点顺序相加，处理好进位即可。关于链表，这里用了几个优化的办法：

1. 使用附加头结点可以避免头结点需要单独处理的情况。

2. 进位用一个bool类型来表示，要注意处理的特殊情况包括：链表长度不一致，链表两数相加有进位，在最后一位有进位

代码如下，运行时间40ms：

/**
* 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) {
// 特殊情况处理
if(l1 == NULL && l2 == NULL)
return NULL;
else if(l1 == NULL)
return l2;
else if(l2 == NULL)
return l1;

// 进位符
bool carry = false;
// 附加头结点
ListNode *head = new ListNode(0), *curr = head;

while(l1 != NULL && l2 != NULL) {
curr->next = addDigit(l1->val, l2->val, carry);
curr = curr->next;
l1 = l1->next;
l2 = l2->next;
}

// 让l1指向未完链表头
if(l1 == NULL)
l1 = l2;

while(carry && l1 != NULL) {
curr->next = addDigit(l1->val, 0, carry);
curr = curr->next;
l1 = l1->next;
}
// 最高位有进位
if(carry)
curr->next = new ListNode(1);

// 复制剩余节点
copyList(curr, l1);

// 删除附加头结点
curr = head;
head = head->next;
delete curr;

return head;
}

void copyList(ListNode *l1, ListNode *l2) {
while(l2 != NULL) {
l1->next = new ListNode(l2->val);
l1 = l1->next;
l2 = l2->next;
}
}

ListNode *addDigit(int val1, int val2, bool &carry) {
int val(val1 + val2);
if(carry) {
++val;
carry = false;
}
if(val >= 10) {
val -= 10;
carry = true;
}
return new ListNode(val);
}
};