leetcode 83 Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

下面是我的解决方案，考虑测试用例：
1，1
1，1，1
1，2，2

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head)
    {
        if(head==NULL)return head;

        ListNode* pre = head;
        ListNode* cur = head->next;

        while(cur!=NULL)
        {
            if(cur->val==pre->val)
            {
                pre->next = pre->next->next;
                cur = cur->next;
                if(cur==NULL)return head;
                //free(cur)没有free是不对的，可能引起内存泄漏;
            }
            else if(cur->val!=pre->val)
            {
                pre = pre->next;
                cur = cur->next;
            }

        }
        return head;

    }
};

c++：

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) return head;

        ListNode cur = head;
        while(cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            }
            else cur = cur.next;
        }
        return head;
    }
}

c语言：

struct ListNode* deleteDuplicates(struct ListNode* head)
{
    if (head) {
    struct ListNode *p = head;
    while (p->next) {
        if (p->val != p->next->val) {
            p = p->next;
        }
        else {
            struct ListNode *tmp = p->next;
            p->next = p->next->next;
            free(tmp);//这块在实际代码中，非常有必要
        }
    }
}

return head;

}

简洁的python解决方案：

class Solution:
    # @param head, a ListNode
    # @return a ListNode
    def deleteDuplicates(self, head):
        node = head
        while node:
            while node.next and node.next.val == node.val:
                node.next = node.next.next

            node = node.next

        return head