力扣链接

方法一:

使用前后两个指针,cur指向当前位置,prev指向前一个位置,通过改变指向和释放结点来删除val

初步代码,还存在问题:

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* removeElements(struct ListNode* head, int val)
{struct ListNode* prev = NULL;struct ListNode* cur = head;while(cur){if(cur->val != val){prev = cur;cur = cur->next;}else{prev->next = cur->next;free(cur);// cur = cur->next;//错误,cur已经被释放,野指针cur = prev->next;}}return head;}

null pointer出现了空指针

通过测试用例代码走读分析问题:

如果第一个就是要删的值,也就是头删,会出现问题

所以这种情况要单独处理

最终代码:

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* removeElements(struct ListNode* head, int val)
{struct ListNode* prev = NULL;struct ListNode* cur = head;while(cur){if(cur->val != val){prev = cur;cur = cur->next;}else{if(prev == NULL){head = cur->next;free(cur);cur = head;}else{prev->next = cur->next;free(cur);//cur = cur->next;//和下一句等价,但是cur已经释放,这句会出现野指针cur = prev->next;}}}return head;//返回一个新的头,不需要用二级指针}

方法二:

把不是val的值尾插到新链表

初步代码:

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* removeElements(struct ListNode* head, int val)
{struct ListNode* newHead= NULL,* tail = NULL;struct ListNode* cur = head;while(cur){if(cur->val != val){//尾插if(tail == NULL){newHead = tail = cur;}else{tail->next = cur;tail = tail->next;}cur = cur->next;}else{struct ListNode* next = cur->next;free(cur);cur = next;}}return newHead;
}

通过走读代码我们发现,当最后一个结点的值为val时,释放节点后前面尾插的结点仍然指向最后一个结点,这里只需要将tail->next置空即可,修改后代码如下:

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* removeElements(struct ListNode* head, int val)
{struct ListNode* newHead= NULL,* tail = NULL;struct ListNode* cur = head;while(cur){if(cur->val != val){//尾插if(tail == NULL){newHead = tail = cur;}else{tail->next = cur;tail = tail->next;}cur = cur->next;}else{struct ListNode* next = cur->next;free(cur);cur = next;}}tail->next = NULL;return newHead;
}

但是代码仍然存在错误,运行如下:

显而易见,需要考虑链表为空的情况

改进后代码:

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* removeElements(struct ListNode* head, int val)
{if(head== NULL){return NULL;}struct ListNode* newHead= NULL,* tail = NULL;struct ListNode* cur = head;while(cur){if(cur->val != val){//尾插if(tail == NULL){newHead = tail = cur;}else{tail->next = cur;tail = tail->next;}cur = cur->next;}else{struct ListNode* next = cur->next;free(cur);cur = next;}}tail->next = NULL;return newHead;
}

报错:

这时代码直接指向最后一个else,此时tail为空,tail->next不合理,所以干脆前面不进行判断,而在后面对tail进行判断

最终代码如下:

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* removeElements(struct ListNode* head, int val)
{struct ListNode* newHead= NULL,* tail = NULL;struct ListNode* cur = head;while(cur){if(cur->val != val){//尾插if(tail == NULL){newHead = tail = cur;}else{tail->next = cur;tail = tail->next;}cur = cur->next;}else{struct ListNode* next = cur->next;free(cur);cur = next;}}if(tail){tail->next = NULL;}return newHead;
}