题目：
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.
代码：
/**
* 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 dummy(-);
ListNode *p = &dummy;
while ( l1 && l2 )
{
if ( l1->val<l2->val )
{
p->next = l1;
l1 = l1->next;
}
else
{
p->next = l2;
l2 = l2->next;
}
p = p->next;
}
p->next = l1 ? l1 : l2;
return dummy.next;
}
};
tips：
无。
========================================
第二次过这道题，l1或l2为空之后，不用一个一个链接了，可以直接补上剩下的linked list。
/**
* 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 head();
ListNode* p = &head;
while ( l1 && l2 )
{
if ( l1->val < l2->val )
{
p->next = l1;
l1 = l1->next;
}
else
{
p->next = l2;
l2 = l2->next;
}
p = p->next;
}
p->next = l1 ? l1 : l2;
return head.next;
}
};