第一天 Add Two Numbers（链表加法）

这是leetcode第二题，对链表的知识基础有一定要求。暂时自己写不出来，先完全解析别人的代码。

1. 先排除极端情况，简化后续

<span style="font-size:14px;"> if(l1 == NULL && l2) return l2;
if(l1 && l2 == NULL) return l1;
if(l1 == NULL && l2 == NULL) return NULL;</span>

2. p1,p2为了避免对原输入的影响；head用于最后的输出；c是进位问题；r是随着链表深入下去；bUseList2是决定r沿着l1或者l2进行
   由于l1,l2有完整的链表体系，所以沿着l1,l2的next可以避免新生成对象的麻烦。换而言之，借用现有的“变量”l1或者l2，修改成为result

<span style="font-size:14px;"> ListNode * p1 = l1;
ListNode * p2 = l2;
ListNode *r = l1; // r indicates the node space we will use for one digit of the result.
ListNode *head = r; // head is the result list.
bool bUseList2 = false; // Use this to indicate whether we should start to use l2 as resource pool.
int c = 0;</span>

3. while语句表达了三种情况，即l1,l2耗尽以及没有进位的情况（考虑：怎么处理l1,l2同时耗尽且进位的情况）；r = r->next正是借用l1,l2的巧妙之处。
<span style="font-size:14px;">while(p1 || p2 || c){
int v = c;
if(p1) v += p1->val;
if(p2) v += p2->val;

c = v >= 10? 1: 0;
r->val = v % 10;

if(p1) p1 = p1->next;
if(p2) p2 = p2->next;

// If we haven't started to used l2, and we have reached the tail of l1,
// switch l2 for the next result digit.
if(bUseList2 == false && r->next == NULL){
r->next = l2;
bUseList2 = true;
}

if(p1 || p2 || c)
r = r->next;
}</span>

最后把r->next用NULL堵上即可。

完整代码如下：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
// first take care of the empty list cases.
if(l1 == NULL && l2) return l2;
if(l1 && l2 == NULL) return l1;
if(l1 == NULL && l2 == NULL) return NULL;

ListNode * p1 = l1;
ListNode * p2 = l2;
ListNode *r = l1; // r indicates the node space we will use for one digit of the result.
ListNode *head = r; // head is the result list.
bool bUseList2 = false; // Use this to indicate whether we should start to use l2 as resource pool.
int c = 0;
while(p1 || p2 || c){
int v = c;
if(p1) v += p1->val;
if(p2) v += p2->val;

c = v >= 10? 1: 0;
r->val = v % 10;

if(p1) p1 = p1->next;
if(p2) p2 = p2->next;

// If we haven't started to used l2, and we have reached the tail of l1,
// switch l2 for the next result digit.
if(bUseList2 == false && r->next == NULL){
r->next = l2;
bUseList2 = true;
}

if(p1 || p2 || c)
r = r->next;
}
// Set the tail of the result list to NULL.
r->next = NULL;

return head;
}