struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: bool hasCycle(ListNode *head) { if(head == NULL) return false; ListNode* walker = head; ListNode* runner = head; while(runner->next != NULL && walker->next != NULL...
针对你提出的runtime error: member access within null pointer of type 'struct listnode'错误,我将按照提供的tips进行分点回答,并附上可能的代码片段以佐证我的建议。 1. 确认错误原因 这个错误表明在代码中尝试访问一个空指针(NULL)指向的struct listnode结构体的成员。在C语言中,如果指针未被正确初始化或分配...
struct ListNode* deleteDuplicates(struct ListNode* head ) { if (head == NULL || head-gt;next == NULL) { return head; _牛客网_牛客在手,offer不愁
这是修正后的代码: structListNode*mergelists(structListNode*list1,structListNode*list2) { structListNode*p=(structListNode*)malloc(sizeof(structListNode)); p->next=NULL; structListNode*head=p; while(list1&&list2) { if(list1->data>list2->data) { p->next=list2; list2=list2->next; } e...
(struct ListNode* head, int m, int n ) { // write code here struct ListNode*prev=NULL; struct ListNode*next=NULL; struct ListNode*phead=head; struct ListNode*reverseBegin=NULL; struct ListNode*reverseEnd=NULL;//define struct,define as NULL int i=1; if(NULL==head||NULL==head->next||...
(head->next); // head->next->next=head; // head->next=NULL; // return newnode; } //编写计算链表长度的函数 int get_len(struct ListNode* head) { int len=0; while(head) { len++; head=head->next; } return len; } struct ListNode* addInList(struct ListNode* head1, struct ...
returnhead; } // 打印链表 voidprintList(ListNode*head){ if(head==nullptr||head->next==nullptr){ return; } ListNode*cur=head->next; while(cur!=nullptr){ cout<<cur->data<<" "; cur=cur->next; } cout<<endl; } // 移动元素位置 ...
做单链表反转题目,报错:member access within null pointer of type ‘struct ListNode’ 题目链接:LINK 问题描述 我明明在初始化指针时候,已经处理了n2->next情况却依然报错 这个报错提示含义是:大概就是他给你传个空指针的话你的语法是错误的含义。
INIT_HLIST_HEAD(&pid->tasks[type]); upid= pid->numbers + ns->level; spin_lock_irq(&pidmap_lock);if(!(ns->pid_allocated &PIDNS_ADDING))gotoout_unlock;for( ; upid >= pid->numbers; --upid) {/*Make the PID visible to find_pid_ns.*/idr_replace(&upid->ns->idr, pid, u...
struct ListNode* ReverseList(struct ListNode* head ) { if (head==NULL||head-gt;next==NULL) { return head; } struct ListNode_牛客网_牛客在手,offer不愁