删除链表中所有含有val的节点

给你一个链表的头节点 head 和一个整数 val ，请你删除链表中所有满足 Node.val == val 的节点，并返回 新的头节点 。

示例 1：

输入：head = [1,2,6,3,4,5,6], val = 6
输出：[1,2,3,4,5]
思路1：遍历查找，找到一个删一个

代码：

#define _CRT_SECURE_NO_WARNINGS
#include
#include

struct ListNode
{
	int val;
	struct ListNode* next;
};

struct ListNode* removeElements(struct ListNode* head, int val)

{
    struct ListNode* cur = head;
    struct ListNode* pre = NULL;
    while (cur != NULL)
    {
        if (cur->val == val)
        {
            if (cur == head)
            {
                head = cur->next;
                free(cur);
                cur = head;
            }
            else
            {
                pre->next = cur->next;
                free(cur);
                cur = pre->next;
            }
        }
        else
        {
            pre = cur;
            cur = cur->next;
        }

    }
    return head;
}
int main()
{
	struct ListNode* n1= (struct ListNode* )malloc(sizeof(struct ListNode));
	struct ListNode* n2 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode* n3 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode* n4 = (struct ListNode*)malloc(sizeof(struct ListNode));

	n1->val = 7;
	n2->val = 6;
	n3->val = 7;
	n4->val = 6;

	n1->next = n2;
	n2->next = n3;
	n3->next = n4;
	n4->next = NULL;

	struct ListNode* head= removeElements(n1,7);

    struct ListNode* cur = head;
    while (cur)
    {
        printf("%d->", cur->val);
        cur = cur->next;
    }
    printf("NULL");

	return 0;
}

思路而，重新定义一个头节点指针=NULL；遍历链表把不等于val的节点移到新的头指针节点处，新城新的链表


代码:

struct ListNode* removeElements1(struct ListNode* head, int val)
{
    struct ListNode* cur = head;
    struct ListNode* newhead = NULL;
    struct ListNode* tail = NULL;
    while (cur)
    {
        if (cur->val == val)
        {
            struct ListNode* pre = cur;
            cur = cur->next;
            free(pre);

        }
        else
        {
            if (tail == NULL)
            {
                newhead = tail = cur;
            }
            else
            {
                tail->next = cur;
                tail = tail->next;

            }
            cur = cur->next;
        }
        if(tail)
        tail->next = NULL;
    }
}