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* deleteDuplicates(struct ListNode* head ) { if (head == NULL || head->next == NULL) { return head; } // write code here typedef struct ListNode ListNode; ListNode* pre = (ListNode*)malloc(sizeof(ListNode)); ListNode* pre_pre = (ListNode*)malloc(sizeof(ListNode)); Li...
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...
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...
typedef struct list { int data; struct list *next; } ListNode; 2. 实现一个函数,用于遍历并打印单链表中的所有元素 接下来,我们需要实现一个函数来遍历并打印单链表中的所有元素。这个函数将从头节点开始,沿着链表遍历,直到遇到空指针。 c void printList(ListNode *head) { ListNode *current = head; ...
(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||...
};intmain(){structlistNode*head=NULL,*node=NULL,*pListEnd=NULL;node=(structlistNode*)malloc(...
classListNode{varvalue:Int64varnext:ListNode?// 可选引用,允许nullinit(value: Int64, next: ListNode? = nil) { self.value= value self.next= next } }// 构建链表:1 -> 2 -> 3letnode3 =ListNode(value:3)letnode2 =ListNode(value:2,next: node3)lethead =ListNode(value:1,next: node2) ...
let node2 = ListNode(value: 2, next: node3) let head = ListNode(value: 1, next: node2) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 复制 枚举+类组合实现逻辑递归 通过枚举成员包裹类实例,间接实现层级结构。 enum Tree {
(head1); struct ListNode* q=reverse(head2); struct ListNode* ans=p,*pre=NULL; int in=0; while (p&&q) { int sum=p->val+q->val+in; p->val=sum%10; in=sum/10; if(p->next==NULL)pre=p; p=p->next; q=q->next; } if(q){ p=pre; p->next=q; p=p->next; } while...