题解 | #合并两个排序的链表#

构建头节点一定要初始化 （归并排序）的应用

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        if(!pHead1) return pHead2;
        if(!pHead2) return pHead1;
        ListNode* vhead=new ListNode(0);;
        ListNode* cur;
        if(pHead1->val<=pHead2->val){
            vhead->next=pHead1;
            pHead1=pHead1->next;
        }
        else {vhead->next=pHead2;
             pHead2=pHead2->next;
             }
        cur=vhead->next;
        while(pHead1||pHead2){
            if(!pHead1){
                cur->next=pHead2;
                break;
            }
            if(!pHead2){
                cur->next=pHead1;
                break;
            }
            if(pHead1->val<=pHead2->val){
                cur->next=pHead1;
                cur=cur->next;
                pHead1=pHead1->next;
            }else{
                cur->next=pHead2;
                cur=cur->next;
                pHead2=pHead2->next;
            }
        }
        return vhead->next;
    }
};