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...
int main() { // 示例使用 struct ListNode *head = NULL; // 假设head已经被初始化为一个包含一些节点的链表 int m = 10; // 要删除的值 head = deletem(head, m); // 打印删除后的链表(假设有printlist函数) // printlist(head); return 0; } 注意事项 在实际使用时,你需要确保链表节点的内...
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
(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 ...
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。
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 ...
cpp里的class就是由struct衍生而来，相当于struct + 函数，再添一些新特性。用class描述人就更全面了，...
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 ) { // 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; } } ...