移除链表元素 ＜难度系数＞

1.移除链表元素 ＜难度系数＞

题述：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]

示例2：
输入：head = [ ], val = 1
输出：[ ]

示例3：
输入：head = [7,7,7,7], val = 7
输出：[ ]

🧷 平台：Visual studio 2017 && windows

核心思想：
思路1，双指针，cur找到val值所在的节点，prev记录前一个位置，依次删除，细节请看下图：

思路2，新链表，把链表中所有不是val的值尾插至新链表，删除val的节点

leetcode原题

#include<stdio.h>#include<stdlib.h>typedef int SLTDataType;struct ListNode{	int val;	struct ListNode *next;};//思路1struct ListNode* removeElements1(struct ListNode* head, int val){	struct ListNode * cur = head;	struct ListNode * prev = NULL;	while (cur)//判断地址	{		//1、相等的情况		if (cur->val == val)		{			if (cur == head)//1.1、第1个节点等于val时			{				head = head->next;				free(cur);				cur = head;			}			else//1.2、其它节点等于val时			{				prev->next = cur->next;				free(cur);				cur = prev->next;			}		}		//2、不相等的情况		else		{			prev = cur;			cur = cur->next;		}	}	return head;}//思路2struct ListNode* removeElements2(struct ListNode* head, int val){		if(head == NULL)		return NULL;	//创建新节点	struct ListNode* newhead = NULL, *tail = NULL;	struct ListNode* cur = head;	while(cur)	{		struct ListNode* next = cur->next;		if(cur->val == val)		{			free(cur);		}		else		{			if(tail == NULL)			{				newhead = tail = cur;				}			else			{				tail->next = cur;					tail = cur;			}		}		cur = next;	}	if(tail)		tail->next = NULL;	return newhead;}int main(){	struct ListNode* n1 = (struct ListNode*)malloc(sizeof(struct ListNode));	n1->val = 1;	struct ListNode* n2 = (struct ListNode*)malloc(sizeof(struct ListNode));	n2->val = 2;	struct ListNode* n3 = (struct ListNode*)malloc(sizeof(struct ListNode));	n3->val = 6;	struct ListNode* n4 = (struct ListNode*)malloc(sizeof(struct ListNode));	n4->val = 3;	struct ListNode* n5 = (struct ListNode*)malloc(sizeof(struct ListNode));	n5->val = 4;	struct ListNode* n6 = (struct ListNode*)malloc(sizeof(struct ListNode));	n6->val = 5;	struct ListNode* n7 = (struct ListNode*)malloc(sizeof(struct ListNode));	n7->val = 6;	n1->next = n2;	n2->next = n3;	n3->next = n4;	n4->next = n5;	n5->next = n6;	n6->next = n7;	n7->next = NULL;	removeElements1(n1, 6);	removeElements2(n1, 6);	return 0;}