LeetCode： Remove Linked List Elements

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6
Return: 1 –> 2 –> 3 –> 4 –> 5

解析：

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* newhead = new ListNode(-1);
        newhead->next = head;
        ListNode* pre = newhead;
        ListNode* cur = head;
        while(cur != NULL)
        {
            if(cur->val == val)
            {
                cur = cur->next;
                pre->next = cur;
            }
            else
            {
                pre = cur;
                cur = cur->next;
            }
        }
        return newhead->next;
    }
};

方法二：

void removeHelper(ListNode *&head, int val)
{
    if (head == nullptr)
    {
        return;
    }
    else if (head->val == val)
    {
        head = head->next;
    }
    else
    {
        removeHelper(head->next, val);
    }
}

方法三（有问题一直不能AC）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        if (!head)
            return NULL;
        if (head->next == NULL)
            return head;
        ListNode *p = head->next;
        ListNode *pre = head;
        while (p)
        {
            if (p->val == val)
            {
                ListNode *r = p->next;
                pre->next = r;
                free(p);
                p = r;
            }
            else
            {
                pre = p;
                p = p->next;
            }

        }
        return head;
    }
};

方法四：

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode *p = head;
        ListNode *pre = head;
        while (p)
        {
            if (p->val == val)
            {
                if(p==head)
                    pre = head = p->next;
                else
                    pre->next = p->next;

                free(p);
                p = pre;
            }
            else
            {
                pre = p;
                p = p->next;
            }

        }
        return head;
    }
};