可以创建一个新链表记录答案：

/**
 * 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* list1, ListNode* list2) {
        ListNode* result=new ListNode();
        ListNode* now=result;
        while(list1||list2){
            if(list2==nullptr||list1&&list1->val<list2->val){
                now->next=list1;
                list1=list1->next;
            }
            else if(list1==nullptr||list2&&list1->val>=list2->val){
                now->next=list2;
                list2=list2->next;
            }
            now=now->next;
        }
        return result->next;
    }
};

也可以用递归直接在原有链表上创建答案：

/**
 * 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* list1, ListNode* list2) {
        if(list1==nullptr) return list2;
        if(list2==nullptr) return list1;
        if(list1->val<list2->val){
            list1->next=mergeTwoLists(list1->next,list2);
            return list1;
        }
        else{
            list2->next=mergeTwoLists(list1,list2->next);
            return list2;
        }
    }
};