【牛客题霸题解】删除有序链表中重复出现的元素

从链表的第一个节点开始遍历，如果当前的结点的值等于下一个结点的值，就说明这个结点是要删除的。然后看一下后面有多少个结点等于这个值，一起删除掉。
因为第一个结点也有可能被删除，所以在第一个结点前加一个头节点。

图中深色的部分需要删除

c++

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head == NULL){return head;}
        ListNode* front = new ListNode(-1);
        front->next = head;
        ListNode* now = front->next;
        ListNode* pre = front;
        ListNode* next = NULL;
        while(now!=NULL)
        {
            if(now->next!=NULL&& now->val == now->next->val)
            {
                ListNode* Dstart = pre;
                pre = now;
                now = now->next;
                while(now!=NULL&&now->val==pre->val)
                {
                    delete pre;
                    pre = now;
                    now = now->next;
                }
                Dstart->next = now;
                pre = Dstart;
                continue;
            }
            pre = now;
            now = now->next;
        }
        return front->next;
    }
};

java

import java.util.*;
public class Solution {
    public ListNode deleteDuplicates (ListNode head) {
        if(head == null){return head;}
        ListNode front = new ListNode(-1);
        front.next = head;
        ListNode now = front.next;
        ListNode pre = front;
        ListNode next = null;
        while(now!=null)
        {
            if(now.next!=null && now.val == now.next.val)
            {
                ListNode Dstart = pre;
                pre = now;
                now = now.next;
                while(now!=null&&now.val==pre.val)
                {
                    pre = now;
                    now = now.next;
                }
                Dstart.next = now;
                pre = Dstart;
                continue;
            }
            pre = now;
            now = now.next;
        }
        return front.next;
    }
}

python

class Solution:
    def deleteDuplicates(self , head ):
        if head == None:
            return head
        front = ListNode(-1)
        front.next = head
        now = front.next
        pre = front
        next = None
        while now!=None:
            if now.next!=None and now.val == now.next.val:
                Dstart = pre
                pre = now
                now = now.next
                while now!=None and now.val==pre.val:
                    pre = now
                    now = now.next
                Dstart.next = now
                pre = Dstart
                continue
            pre = now
            now = now.next
        return front.next