題目：

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.

合併兩個排序的連結串列，並返回一個新的連結串列。新連結串列應由兩個連結串列拼接而成。

/**
 * 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) 
    {
        //l1為空，則返回l2，l2為空，則返回l1
        if (l1 == NULL) return l2;
        if (l2 == NULL) return l1;
        
        //q結點記錄連結串列的起始結點
        ListNode q(-1);
        ListNode *p=&q;
        
        //迴圈判斷l1和l2大小，小的連結到p後，迴圈結束條件是l1和l2之間有一個先遍歷完
        while(l1 != NULL && l2 != NULL)
        {
        	if(l1->val <= l2->val)
        	{
        		p->next = l1;
        		p = p->next;
        		l1 = l1->next;
        	}
        	else
        	{
        		p->next = l2;
        		p = p->next;
        		l2 = l2->next;
        	}
        }
        //如果l2先遍歷完，把l1接到p後即可
        if(l1)
        {
        	p->next = l1;
        }
        if(l2)
        {
        	p->next = l2;
        }
        //返回連線後連結串列的首結點，不是q，q不屬於l1和l2，而是q->next
        return q.next;
    }
};
 

遞迴解法：

/**
 * 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 *p = NULL;
        if(l1->val < l2->val)
        {
            p = l1;
            p->next = mergeTwoLists(l1->next , l2);
        }
        else
        {
            p = l2;
            p->next = mergeTwoLists(l1 , l2->next);
        }
        return p;
    }
};