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. 思路.首先判断是否有空链表,如果有,则直接返回另一个链表,如果没有,则开始比较两个链表的当前节点,返回较小的元素作为前驱,并且指针向后移动一位,再进行比较,如此循环,知道一个链表的next指向NULL,将另一个链表的…