如何解决类型的空指针内的成员访问,C 编程是回文
我正在尝试使用非递归解决方案解决 LeetCode 上的 isPalindrome() 问题,当我使用 VSCode 运行此代码时,它会运行并给我正确的输出,但是当我在 LeetCode 编译器中运行它时,它给了我下面提到的错误.
你能帮我解决这个问题吗?对我的代码是否有任何修改可以帮助我以更好的方式解决问题?
bool isPalindrome(struct ListNode* head){
struct ListNode * NewHead,*MidList,*EndList;
NewHead = EndList = MidList = head;
struct ListNode *ptr_SecondHalf,*ptr_FirstHalf;
while (EndList->next != NULL)
{
EndList = EndList->next->next;
MidList = MidList->next;
if (EndList->next == NULL || EndList->next->next == NULL){
//ODD
if (EndList->next == NULL){
ptr_SecondHalf = MidList->next;
break;
}
//EVEN
if (EndList->next->next == NULL){
ptr_SecondHalf = MidList->next;
break;
}
}
}
MidList->next = NULL;
ptr_FirstHalf = head;
// Reverse SecondHalf
struct ListNode* ptr_SecondHalf_Reversed = NULL;
struct ListNode* current = ptr_SecondHalf;
struct ListNode* next = NULL;
while (current != NULL) {
// Store next
next = current->next;
// Reverse current node's pointer
current->next = ptr_SecondHalf_Reversed;
// Move pointers one position ahead.
ptr_SecondHalf_Reversed = current;
current = next;
}
while (ptr_FirstHalf->next != NULL && ptr_SecondHalf_Reversed != NULL){
if (ptr_FirstHalf->val == ptr_SecondHalf_Reversed->val){
if (ptr_FirstHalf->next->next != NULL || ptr_SecondHalf_Reversed ->next != NULL){
ptr_FirstHalf = ptr_FirstHalf->next;
ptr_SecondHalf_Reversed = ptr_SecondHalf_Reversed ->next ;
}
else
{
break;
}
}else{
printf("false \n");
return false;
}
}
printf("true \n");
return true;
}
第 12 行:字符 21:运行时错误:在“struct ListNode”[solution.c] 类型的空指针内访问成员
解决方法
您的解决方案思路没有错,但是太难实现,没有任何错误,我建议我的解决方案或建议编写一个返回反向列表的方法,然后比较这两个。
bool isPalindromeRec(struct ListNode* left,struct ListNode* right) {
if(right == NULL) return true;
if(isPalindromeRec(left,right->next) && left->val == right->val) {
left = left->next;
return true;
}
return false;
}
bool isPalindrome(struct ListNode* head){
if(head == NULL) return true;
struct ListNode * left = head;
struct ListNode * right = head;
bool ans = isPalindromeRec(left,right);
// happy now Vlad from Moscow?
if(ans) printf("true\n");
else printf("false\n");
return ans;
}
我刚刚提出的这个解决方案应该是正确的 - 它只是从左到右比较每个节点,但我认为这不是会发生很大变化的时间。 编辑:
bool isPalindrome(struct ListNode *head) {
if(head == NULL|| head->next== NULL)
return true;
//find list center
struct ListNode *fast = head;
struct ListNode *slow = head;
while(fast->next != NULL && fast->next->next != NULL){
fast = fast->next->next;
slow = slow->next;
}
struct ListNode *secondHead = slow->next;
slow->next = NULL;
//reverse second part of the list
struct ListNode *p1 = secondHead;
struct ListNode *p2 = p1->next;
while(p1 != NULL && p2 != NULL){
struct ListNode *temp = p2->next;
p2->next = p1;
p1 = p2;
p2 = temp;
}
secondHead->next = NULL;
//compare two sublists now
struct ListNode *p = (p2 == NULL?p1:p2);
struct ListNode *q = head;
while(p!=NULL){
if(p->val != q->val)
return false;
p = p->next;
q = q->next;
}
return true;
}
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。