递归版：确实简单很多，但是还需要练习。

class Solution {  
public:  
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {  
        if (l1 == NULL) {  
            return l2;  
        }  
        if (l2 == NULL) {  
            return l1;  
        }  
        if (l1->val <= l2->val) {  
            l1->next = mergeTwoLists(l1->next, l2);  
            return l1;  
        } else {  
            l2->next = mergeTwoLists(l1, l2->next);  
            return l2;  
        }  
    }  
};  

正常拼接版：

/**
 * 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 *sum=new ListNode(0);
        ListNode *cur=sum;
        if(l1==NULL) return l2;
        if(l2==NULL) return l1;
        while(l1!=NULL&&l2!=NULL)
        {
            if(l1->val<=l2->val)
            {
                cur=(cur->next=new ListNode(l1->val));
                //printf("%d %d\n",l1->val,l2->val);
                l1=l1->next;
            }
            else{
                cur=(cur->next=new ListNode(l2->val));
                //printf("%d %d\n",l1->val,l2->val);
                l2=l2->next;
            }
        }
        while(l1!=NULL)
        {
             cur=(cur->next=new ListNode(l1->val));
             l1=l1->next;
        }
        while(l2!=NULL)
        {
            cur=(cur->next=new ListNode(l2->val));
            l2=l2->next;
        }
        return sum->next;
    }
};