C++ 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2020-11-23

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* deleteDuplicates(ListNode* head) {
        // write code here
        /*if (!head || !head->next) return head;
        if(head->val==head->next->val){
            return nullptr;
        }
         ListNode *start = head;
         while (start && start->next) {
            if (start->val == start->next->val) {
                 ListNode *tmp = start->next;
                 start->next = start->next->next;
                 delete tmp;
             } else start = start->next;
         }
         return head;*/
        //https://blog.csdn.net/No_one_z/article/details/109634379
        if(head==nullptr || head->next==nullptr)return head;
        ListNode* newHead = new ListNode(1);
        newHead->next=head;
        ListNode*pre = newHead;
        ListNode*cur = head;
        int count=0;
        while(cur && cur->next){
            if(cur->val==cur->next->val){
                cur=cur->next;
                count++;
            }else{
                if(count > 0){
                    pre->next=cur->next;
                    cur=cur->next;
                    count =0;
                }else{
                    pre=pre->next;
                    cur=cur->next;
                }
            }
        }
        if(count > 0){
            pre->next=cur->next;
        }
        return newHead->next;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2020-11-22

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* deleteDuplicates(ListNode* head) {
        // write code here
        if(head==nullptr||head->next==nullptr)
            return head;
        ListNode newhead(0);
        newhead.next=head;
        ListNode *pnode=&newhead;
        while(pnode->next)
        {
            ListNode *ptemp=pnode->next;
            bool flag=false;
            while(ptemp&&ptemp->next&&ptemp->val==ptemp->next->val)
            {
                ptemp=ptemp->next;
                flag=true;
            }
            if(flag)
                pnode->next=ptemp->next;
            else
                pnode=pnode->next;
        }
        return newhead.next;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 332KB, 提交时间: 2021-05-24

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* deleteDuplicates(ListNode* head) {
        // write code here
        ListNode *dummy = new ListNode(0);//省去判断链表头重复的情况
        dummy->next = head;
        ListNode *cur=head,*pre=dummy,*q;//记录前一个没重复的节点和当前节点
        
        if(head==NULL||head->next==NULL){
            return head;
        }
        
        while(cur&&cur->next){//后面元素少于2个则不会重复
            bool repeat = 0;//是否重复标志位
            while(cur->next && cur->next->val==pre->next->val){//存在下一节点并且下一节点的值与pre后面的值相同时（即接下来要判断是否重复的节点值）
                repeat = 1;
                cur = cur->next;
            }
            if(repeat){//当有重复时前一个没重复的节点指向末尾重复节点的下一位（可能为下一位）
                pre->next = cur->next;
            }
            else{//否则没重复节点前移
                pre = cur;
            }
            cur = cur->next;
        }
        return dummy->next;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 332KB, 提交时间: 2021-05-22

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* deleteDuplicates(ListNode* head) {
        // write code here
        ListNode* dummy=new ListNode(233);
        dummy->next=head;
        ListNode* p=head,*pre=dummy,*q;
        if(head==NULL||head->next==NULL) return head;
        while(p&&p->next)
        {
            q=p;
            while(q&&q->val==p->val) q=q->next;
            if(q==NULL)
              pre->next=NULL;
            if(q==p->next)
            {
                pre->next=p;
                pre=p;
            }
            p=q;
        }
        if(p&&p->next==NULL)
            pre->next=p;
        return dummy->next;
        
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 332KB, 提交时间: 2020-11-21

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* deleteDuplicates(ListNode* head) {
        // write code here
        if(head==nullptr || head->next==nullptr)
            return head;
        ListNode* dummy = new ListNode(0);
        dummy->next=head;
        ListNode* p=head;
        ListNode* pre=dummy;
        while(p!=nullptr && p->next!=nullptr)
        {
            if(p->val==p->next->val)
            {
                ListNode* Next=p;
                while(Next->next!=nullptr && (Next->val==Next->next->val))
                    Next=Next->next;
                pre->next=Next->next;
                while(p!=pre->next)
                {
                    Next=p->next;
                    delete p;
                    p=Next;
                }
            }
            else{
                p=p->next;
                pre=pre->next;
            }
        }
        return dummy->next;
    }
};