要求：空间复杂度 O(1) ，时间复杂度 O(n) 。

方法1：新链表合并，抄自牛客题解。比较直观。

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        auto vhead=new ListNode(-1); //设置一个哨兵指针
        ListNode *newhead=vhead;
        while(pHead1&&pHead2)
        {
            if(pHead1->val <= pHead2->val){
                newhead->next = pHead1;
                pHead1 = pHead1->next;
            }else{
                newhead->next = pHead2;
                pHead2 = pHead2->next;
            }
            newhead = newhead->next;
        }
        if(pHead2==nullptr) 
            newhead->next=pHead1;
        if(pHead1==nullptr)
            newhead->next=pHead2;
        return vhead->next;
    }
};

方法2：递归。时间复杂度：O(m+n)；空间复杂度：O(m+n)。

class Solution {
public:
 ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
 {
     if (!pHead1) return pHead2;
     if (!pHead2) return pHead1;
     if (pHead1->val <= pHead2->val) {
         pHead1->next = Merge(pHead1->next, pHead2);
         return pHead1;
     }
     else {
         pHead2->next = Merge(pHead1, pHead2->next);
         return pHead2;
     }
 }
};