[Leetcode]删除链表中等于val 的所有结点

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;
}