203. 移除链表元素
正确运行的代码如下：

truct ListNode* removeElements(struct ListNode* head, int val){while (head && head->val == val){head = head->next;}struct ListNode * cur = head;while (cur && cur->next){// printf("cur->val = %d\n", cur->val);while (cur->next && cur->next->val == val){cur->next = cur->next->next;}cur = cur ->next;}return head;
}

或者通过创建一个虚拟头节点。

struct ListNode* removeElements(struct ListNode* head, int val){struct ListNode * dummyHead = (struct ListNode *)malloc(sizeof(struct ListNode));dummyHead->next = head;struct ListNode * cur = dummyHead;while (cur->next){if (cur->next->val == val){cur->next = cur->next->next;}else {cur = cur->next;}}return dummyHead->next;
}

以下是错误代码，不知道为什么超出时间限制。我通过打印运行结果，初步认为是无法判断链表结束。

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* removeElements(struct ListNode* head, int val){struct ListNode * pHead = NULL, * p = head, * q;while (p && p->val == val){p = p->next;}pHead = p;q = pHead;if (q){while (p){if (p->val != val){q->next = p;q = q->next;}p = p->next;}}return pHead;
}

707. 设计链表
这道题目包含了链表的基本操作，按照代码随想录的视频，设置了虚拟头节点来进行操作。

区分相关概念：
首节点：链表的第一个有效节点。
尾节点：最后一个有效节点。
头节点：链表的第一个有效节点之前的那个节点，头节点并不存放有效数据。加头节点的目的主要是为了方便对链表的操作。
头指针：指向头节点的指针变量。
尾指针：指向尾节点的指针变量。

为什么在遍历链表的时候要定义一个指针来遍历，而不是直接操作头指针？
这是因为我们要操作完链表之后要返回头节点，如果直接操作头节点的话，那头节点的值都被修改了，那么如何返回链表的头节点呢？所以要定义一个临时指针指向头节点进行遍历。

对于这道题目而言，一定要明白第n个节点在哪里，以及如何对链表进行插入数据。
注意链表中的所有节点下标从 0 开始。
正确运行的代码如下。

typedef struct {int val;struct MyLinkedList * next;
} MyLinkedList;MyLinkedList* myLinkedListCreate() {MyLinkedList * dummyhead = (MyLinkedList *) malloc (sizeof(MyLinkedList));dummyhead->val = 0;dummyhead->next = NULL;return dummyhead;
}int myLinkedListGet(MyLinkedList* obj, int index) {MyLinkedList * cur = obj->next;while (index-- && cur){cur = cur->next;}if (cur){return cur->val;}return -1;
}void myLinkedListAddAtHead(MyLinkedList* obj, int val) {MyLinkedList * q = (MyLinkedList *) malloc (sizeof(MyLinkedList));q->val = val;q->next = obj->next;obj->next = q;
}void myLinkedListAddAtTail(MyLinkedList* obj, int val) {MyLinkedList * q = (MyLinkedList *) malloc (sizeof(MyLinkedList));q->val = val;q->next = NULL;MyLinkedList * cur = obj;while (cur->next){cur = cur->next;}cur->next = q;
}void myLinkedListAddAtIndex(MyLinkedList* obj, int index, int val) {MyLinkedList * cur = obj;while (index-- && cur){cur = cur->next;}MyLinkedList * q = (MyLinkedList *) malloc (sizeof(MyLinkedList));q->val = val;q->next = NULL;if (cur){q->next = cur->next;cur->next = q;}
}void myLinkedListDeleteAtIndex(MyLinkedList* obj, int index) {MyLinkedList * cur = obj;MyLinkedList * q;while (index--){cur = cur->next;}if (cur && cur->next){q = cur->next;cur->next = q->next;}}void myLinkedListFree(MyLinkedList* obj) {while (obj){MyLinkedList * q = obj;obj = obj->next;free(q);}
}

以下是错误代码，还没有想明白是为什么。

typedef struct Node{ //定义链表的节点的结构体int val; //数据域struct Node * next; //指针域
}Node, * pNode;typedef struct { //定义链表的结构体pNode dummyhead; //头指针int len; //链表长度
} MyLinkedList;MyLinkedList* myLinkedListCreate() {MyLinkedList * list = (MyLinkedList *) malloc (sizeof(MyLinkedList));list->dummyhead = NULL;list->len = 0;return list;
}int myLinkedListGet(MyLinkedList* obj, int index) {if (index < 0 || index >= obj->len){return -1;}pNode cur = obj->dummyhead;while (index--){cur = cur->next;}return cur->val;
}void myLinkedListAddAtHead(MyLinkedList* obj, int val) {pNode q = (pNode) malloc (sizeof(Node));q->val = val;q->next = NULL;pNode cur = obj->dummyhead;if (cur){q->next = cur->next;cur->next = q;obj->len++;}
}void myLinkedListAddAtTail(MyLinkedList* obj, int val) {pNode q = (pNode) malloc (sizeof(Node));q->val = val;q->next = NULL;pNode cur = obj->dummyhead;while (cur && cur->next){cur = cur->next;}if (cur){cur->next = q;obj->len++;}
}void myLinkedListAddAtIndex(MyLinkedList* obj, int index, int val) {if (index < 0 || index > obj->len){return;}else if (index == 0){myLinkedListAddAtHead(obj, val);}else if (index == obj->len){myLinkedListAddAtTail(obj, val);}else {pNode q = (pNode) malloc (sizeof(Node));q->val = val;q->next = NULL;pNode cur = obj->dummyhead;while (index--){cur = cur->next;}q->next = cur->next;cur->next = q;obj->len++;}
}void myLinkedListDeleteAtIndex(MyLinkedList* obj, int index) {if (index >= 0 && index < obj->len){pNode cur = obj->dummyhead;while (index--){cur = cur->next;}cur->next = cur->next->next;obj->len--;}
}void myLinkedListFree(MyLinkedList* obj) {pNode q, cur = obj->dummyhead;while (obj->len--){q = cur->next;free(q);cur = cur->next;}
}/*** Your MyLinkedList struct will be instantiated and called as such:* MyLinkedList* obj = myLinkedListCreate();* int param_1 = myLinkedListGet(obj, index);* myLinkedListAddAtHead(obj, val);* myLinkedListAddAtTail(obj, val);* myLinkedListAddAtIndex(obj, index, val);* myLinkedListDeleteAtIndex(obj, index);* myLinkedListFree(obj);
*/

206. 反转链表
注意掌握双指针解法以及递归写法。

struct ListNode* reverseList(struct ListNode* head){struct ListNode * pre = NULL, * cur = head; //初始化变量while (cur){struct ListNode * tmp = cur->next; //临时变量保存cur->next = pre; //反转pre = cur; //更新变量cur = tmp;}return pre; //返回头节点
}

struct ListNode * reverse(struct ListNode * cur, struct ListNode * pre){if (!cur){return pre;}struct ListNode * tmp = cur->next;cur->next = pre;return reverse(tmp, cur);
}struct ListNode* reverseList(struct ListNode* head){return reverse(head, NULL);
}

继续加油~