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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

就当复习单链表呗

/**
 * 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) {
        struct ListNode * res = NULL;
        struct ListNode * p1 = l1;  //p1  p2 分别用来跟踪 l1,l2; p3永远指向新的链表的最后一个元素
        struct ListNode * p2 = l2;
        struct ListNode * p3 = res;
        while(p1 != NULL && p2 != NULL){
            int x ;
            if( p1->val < p2->val ){
                x = p1->val;
                p1 = p1->next;
            }
            else{
                x = p2->val;
                p2 = p2->next;
            }
            struct ListNode * temp = new ListNode(x);   //新建一个节点
            
            if(p3 == NULL){  //没有头节点就是很烦啊，各种NULL的判断
                res = temp;
                p3 = res;
            }
            else{
                p3->next = temp;
                p3 = p3->next;
            }
        }
        //一个结束了另一个还没有结束 
        if(p1 == NULL && p2 != NULL){
            if(p3 == NULL)    //一个都没加进去- -
                res = p2;
            else
                p3->next = p2;
        }
        else if(p1 != NULL && p2 == NULL){
            if(p3 == NULL)   //同样，一个都没加进去
                res = p1;
            else
                p3->next = p1;
        }
        
        return res;
        
    }
};