第一天 Add Two Numbers（链表加法）

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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; }

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