LootCode21: merge two ordered linked lists

1. Title description

The two merged into a new list in ascending order of ascending list and returns. The new linked list is composed by splicing all the nodes of the given two linked lists.

Example:

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

Two, problem-solving ideas

1. First determine whether there is an empty list

   if(l1==nullptr)
   	return l2;
   if(l2==nullptr)
   	return l1;
   

2. Determine the head node of the merged linked list (the node with a small integer value)

   ListNode* head;
   l1->val>l2->val ? (head=l2,l2=l2->next):(head=l1,l1=l1->next); 
   

3. Connect subsequent nodes

   while(l1!=nullptr&&l2!=nullptr){
         
                     
               l1->val>l2->val ? (p->next=l2,l2=l2->next):(p->next=l1,l1=l1->next);            
               p=p->next;            
           }
   

Three, my code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
    
    
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    
    
        if(l1==nullptr)
            return l2;
        if(l2==nullptr)
            return l1;
        ListNode* head;
        l1->val>l2->val ? (head=l2,l2=l2->next):(head=l1,l1=l1->next);        
        ListNode* p=head;
        while(l1!=nullptr&&l2!=nullptr){
    
                
            l1->val>l2->val ? (p->next=l2,l2=l2->next):(p->next=l1,l1=l1->next);            
            p=p->next;            
        }
        p->next=(l1==nullptr)? l2:l1;        
        return head;
    }
};

Four, test results