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<stdio.h>
#include<stdlib.h>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;}
}