[Leetcode] Merge two sorted lists 合并两已排序的链表

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.

题意：合并两个排好的链表并返回新的链表。

可以使用归并排序，从两链表的表头，取出结点，比较两值，将较小的放在新链表中。如1->3->5->6和2->4->7->8，先将1放入新链表，然后将3和2比较，较小值放入新链表中，从1到3只要pre=pre->next即可。当其中有一条链表被访问完，结束循环，找出该链表，将其接在新链表的后面即可。

/**
 * 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 *newList=new ListNode(-);
        ListNode *pre=newList;
        while(l1&&l2)
        {
            if(l1->val > l2->val)
            {
                pre->next=l2;
                l2=l2->next;
            }
            else
            {
                pre->next=l1;
                l1=l1->next;
            }
            pre=pre->next;
        }
        if(l1)
            pre->next=l1;
        else
            pre->next=l2;

        return newList->next;
    }
};