剑指offer16-合并两个排序链表

Posted 2020-12-06 troubleasy

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

思路：递归和非递归方式

    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
          if(pHead1==NULL) return pHead2;
        if(pHead2==NULL) return pHead1;
        ListNode *pHead3=NULL,*p3=NULL;//这里牛客网测试时，不初始化为null，将无法通过测试
        while(pHead1!=NULL&&pHead2!=NULL)
        {
            if(pHead1->val<=pHead2->val)
            {
                if(pHead3==NULL)
                {
                    p3=pHead1;
                    pHead3=p3;//=pHead1;
                }
                else {
                    p3->next=pHead1;
                    p3=p3->next;
                }
                pHead1=pHead1->next;
             }
             else
             {
                 if(!pHead3)
                 {
                   p3=pHead2;
                   pHead3=p3;
                     
                 }else {
                    p3->next=pHead2;
                    p3=p3->next;
                }
                pHead2=pHead2->next;
             }
        }
        if(pHead1==NULL)
        {
            p3->next=pHead2;
        }
        if(pHead2==NULL){
            p3->next=pHead1;
        }
        return pHead3；
    }

        //递归

    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
          if(pHead1==NULL) return pHead2;
        if(pHead2==NULL) return pHead1;
        ListNode *pHead3;
        if(pHead1->val<=pHead2->val)
        {
            pHead3=pHead1;
            pHead1=pHead1->next;
            
        }else{
            pHead3=pHead2;
            pHead2=pHead2->next;
           
        }
        pHead3->next=Merge(pHead1,pHead2);
        return pHead3;
    }