LeetCode21 Merge Two Sorted Lists

题意：

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. （Easy）

分析：

链表题，注意细节即可。

链表merge和数组merge的不同在于链表不用拷一份出来，倒腾一下指针就可以啦。

注意事项有：

1. dummy node用于输出，head(或者换个名字)用于遍历；

2. l1,l2是不是为空，一个结束遍历之后记得把另一个剩下的加上去。

代码：

 class Solution {
 public:
     ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
         if (l1 == nullptr) {
             return l2;
         }
         if (l2 == nullptr) {
             return l1;
         }
         ListNode dummy();
         ListNode* head = &dummy;
         while (l1 != nullptr && l2 != nullptr) {
             if (l1 -> val < l2 -> val) {
                 head -> next = l1;
                 head = head -> next;
                 l1 = l1 -> next;
             }
             else {
                 head -> next = l2;
                 head = head -> next;
                 l2 = l2 -> next;
             }
         }
         if (l1 != nullptr) {
             head -> next = l1;
         }
         if (l2 != nullptr) {
             head -> next = l2;
         }
         return dummy.next;

     }
 };

优化一下：

head = head -> next是不论if 还是else都要做的，拿出来写；

按本题的写法和ListNode的定义，其实开始不用判断l1，l2是否为空。（但一般来讲还是写上为好...）

代码：

 class Solution {
 public:
     ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
         ListNode dummy();
         ListNode* head = &dummy;
         while (l1 != nullptr && l2 != nullptr) {
             if (l1 -> val < l2 -> val) {
                 head -> next = l1;
                 l1 = l1 -> next;
             }
             else {
                 head -> next = l2;
                 l2 = l2 -> next;
             }
             head = head -> next;
         }
         if (l1 != nullptr) {
             head -> next = l1;
         }
         if (l2 != nullptr) {
             head -> next = l2;
         }
         return dummy.next；
     }
 };

今天七夕，算是半个假期，状态不是很好，水一道链表题练练手啦。明天开始重新步入正轨按照题号刷啦！！！