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) {}
 * };
 本题的关键是增加一个是否要删除当前节点的标识位；
 */
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 == nullptr || head->next == nullptr) return head;
    auto virtual_head = new ListNode(-1);
    virtual_head->next = head;

    auto pre = virtual_head;
    auto cur = virtual_head->next;
    auto delete_cur = false;
    while (cur != nullptr && cur->next != nullptr) {
      if (cur->val == cur->next->val) {
        delete_cur = true;
        auto temp = cur->next;
        cur->next = temp->next;
        delete temp;
      } else {
        if (delete_cur) {
          pre->next = cur->next;
          delete cur;
          cur = pre->next;
          delete_cur = false;
        } else {
          pre = cur;
          cur = cur->next;
        }
      }
    }

    if (delete_cur) {
      pre->next = cur->next;
      delete cur;
    }
    auto new_head = virtual_head->next;
    delete virtual_head;
    return new_head;
  }
};