21. Merge Two Sorted Lists

Problem:

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.

题目很简单，就是直接把两个已经排列的链表进行合并。吐槽一下题目真的不严谨，升序还是降序都没讲。我的做法是新建一个ListNode指针position，然后通过循环比较两个List上的val值，然后链表下一个位置就指向小的值的List上。最后还得判断哪个List先循环完，然后把另一个链上剩下的元素也要合并，而且这些值肯定是有序的了。

Code：

/** * 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) {        ListNode list(INT_MAX);        ListNode *position = &list;                while (l1 != NULL && l2 != NULL) {            if (l1 -> val < l2 -> val) {                position -> next = l1;                l1 = l1 -> next;            } else {                position -> next = l2;                l2 = l2 -> next;            }            position = position -> next;        }        if (l1 == NULL) {            position -> next = l2;        } else {            position -> next = l1;        }        return list.next;    }};