题解 | #删除有序链表中重复的元素-II#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <cstddef>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @return ListNode类
     */
    ListNode* deleteDuplicates(ListNode* head) {
        if (!head) {
            return nullptr;
        }
        ListNode* slow = head;
        ListNode* fast = head->next;
        ListNode* newhead = new ListNode(0);
        newhead->next = head;
        ListNode* cur = newhead;
        if (!fast) {
            return head;
        }
        while(cur->next != NULL && cur->next->next != NULL){
            //遇到相邻两个节点值相同
            if(cur->next->val == cur->next->next->val){
                int temp = cur->next->val;
                //将所有相同的都跳过
                while (cur->next != NULL && cur->next->val == temp)
                    cur->next = cur->next->next;
            }
            else
                cur = cur->next;
        }
       //返回时去掉表头
        return newhead->next;
        // write code here
    }
};

不能用快慢指针，要用cur->next=cur->next->next巧妙地跳过不想要的节点