; struct ListNode *next; }; struct ListNode* mergelists(struct ListNode* list1, struct ListNode* list2) { // 创建一个哑节点作为新链表的头节点,方便后续操作 struct ListNode dummy; dummy.next = NULL; struct ListNode* tail = &dummy; // 遍历两个链表,合并到新的链表中 while (list1 !
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...
我做了两个指针 walker 移动 1 步和 runner 每次迭代移动 2 步。 但随后这段代码给了我一个错误: Line 15: member access within null pointer of type 'struct ListNode' 是什么导致了这个错误?
member access within misaligned address 0x000000000e91 for type 'struct ListNode', which requires 8 byte alignment (ListNode.c) 0x000000000e91: note: pointer points here 错误原因 在程序倒数第6行处申请了一个tmpN指向的结构体ListNode空间,而该结构体中包含next指针,若该节点作为整个链表的最后一个节点,...
反转链表题目:输入一个链表的头结点,反转该链表,并返回反转后链表的头结点。链表结点定义如下:structListNode { intm_nKey; ListNode* m
list = NULL; struct ListNode* body = NULL; for(int i=1; i<listsLen; i++) { listi = lists[i]; if(resultList == NULL) { resultList = listi; continue; } if(listi == NULL) { continue; } while (listi != NULL) { list = resultList; if(listi->val <= resultList->val)...
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; } } ...
{ LinkList *p; if(t!=NULL) { p=t->next; while(p!=NULL) { printf("%c", p->data); p=p->next; } printf("\n"); } } LinkList* GetKthFromTail(LinkList *t, int k) { if(t==NULL) return NULL; LinkList *before, *after; ...
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 7010 第九节(...
struct ListNode* oddEvenList(struct ListNode* head ) { // write code here if(!head || !head->next || !head->next->next) { return head; } struct ListNode* s = head; struct ListNode* d = head->next; struct ListNode* dhead = d; while (s && d) { s->next=d->next; if(!