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...
struct ListNode* ReverseList(struct ListNode* head ) { // write code here struct ListNode* newnode = NULL; if((head == NULL) || (head->next == NULL)) return head; else { newnode = ReverseList(head->next); head->next->next = head; head->next = NULL; return newnode; } } ...
3 -> 4 -> 5 ListNode *head = createNode(1);...以下是一个使用哨兵位头节点逆序单链表的示例代码: // 链表节点的结构体定义 typedef struct ListNode { int val; struct ListNode *next...typedef struct ListNode { int val; struct ListNode *next; } ListNode; // 创建一个新的链表节点 ListNode...
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
structListNode*head=p; while(list1&&list2) { if(list1->data>list2->data) { p->next=list2; list2=list2->next; } else { p->next=list1; list1=list1->next; } p=p->next; } if(list1==NULL) { p->next=list2; }
(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||...
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...
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。
struct ListNode* deleteDuplicates(struct ListNode* head ) { if(!head || !head->next) return head; struct ListNode* result = (struct ListNode*)malloc(sizeof(struct ListNode)); result->next = head; struct ListNode* p = result; struct ListNode* temp = NULL; while (head) { if(head->next...