• 剑指offer16-合并两个排序链表


    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;
        }

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  • 原文地址:https://www.cnblogs.com/trouble-easy/p/12965562.html
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