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) {}
 * };
 */
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) {
        // write code here
        //当头节点为空时返回空
        if (head == nullptr)
            return head;
            //创建两个指针一个在前一个在后
         ListNode* slow = head, *fast = head->next;
         ListNode* p1 = new ListNode(0);
         ListNode* p2 = p1;
         //定义记录数的值
        int t = -1001;
        //当快指针不为空时
        while (fast) {
            //当快慢指针值相同时或慢指针等于记录数
            if (slow->val == fast->val || slow->val == t) {
                //把slow的值赋给t
                t = slow->val;
                slow = slow->next;
                fast = fast->next;
            } else {
                //不相等时把slow连接在p1后面
                p1->next = slow;
                //p1后移
                p1 = p1->next;
                //快慢指针后移
                slow = slow->next;
                fast = fast->next;
                t = p1->val;
            }
        }
        if (slow->val != t) {
            p1->next = slow;
            p1 = p1->next;
        }
        p1->next = nullptr;
        return p2->next;
    }
};