【Leetcode】【Easy】Merge Two Sorted Lists .

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.

解题：

题目比较简单，注意由于表头不确定，因此新建一个结点指向新创建的链表；

解题步骤：

1、判断l1和l2是否有效

2、新建preHead结点，newlist指针指向preHead；

3、循环开始，满足l1和l2都不为空：

　　（1）比较当前l1和l2的值，将较小的穿进newlist中

4、newlist链接l1或l2的剩余部分

5、释放表头，返回newlist；

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
         if (l1 == NULL) return l2;
         if (l2 == NULL) return l1;

         ListNode* prehead = new ListNode();
         ListNode* newlist = prehead;

         while (l1 != NULL && l2 != NULL) {
             if (l1->val > l2->val) {
                 newlist->next = l2;
                 l2 = l2->next;
             } else {
                 newlist->next = l1;
                 l1 = l1->next;
             }
             newlist = newlist->next;
         }

         if (l1 == NULL) newlist->next = l2;
         else newlist->next = l1;

         l1 = prehead->next;
         delete prehead;
         return l1;
     }
 };