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* deleteDuplicates(struct ListNode* head) { // write code here if (head == NULL) { return NULL; } struct ListNode* h3 = NULL; struct ListNode* last = NULL; while (head) { struct ListNode* px = (struct ListNode*)malloc(sizeof(struct ListNode)); px->val = head->val...
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就是在内存空间上的一套相互靠在一起的基本类型的组合。它们作为一个整体可以在内存上搬来搬去，...
struct ListNode { int val; struct ListNode *next; }; 其中,val表单链表节点的值,next指针用于指向链表的下一个节点。 例如,面试比较常考察的“反转单链表”的题目: struct ListNode *reverseList(struct ListNode *head) { if(head == NULL) {
代码语言:txt 复制 void printList(struct ListNode* head) { struct ListNode* current = head; while (current != NULL) { printf("%s ", current->str); current = current->next; } } void updateNode(struct ListNode* node, char* newStr) { strcpy(node->str, newStr); } ...
void printList(ListNode *head) { ListNode *current = head; // 从头节点开始 while (current != NULL) { printf("%d -> ", current->data); // 打印当前节点的数据 current = current->next; // 移动到下一个节点 } printf("NULL "); // 打印链表结束标志 } 3. 在主函数中创建...
(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||...
typedef ListNode * LinkList ; LinkList Leafhead=NULL; Void Inorder (BinTree T) { LinkList s; If(T){ Inorder(T->lchild); If ((!T->lchild)&&(!T->rchild)){ s=(ListNode*)malloc(sizeof(ListNode)); s->data=T->data; s->next=Leafhead; ...
(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...