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) {
        if (!head || !head->next) return head;

        ListNode dummy(0);
        dummy.next = head;
        ListNode* prev = &dummy, *curr = head;

        while (curr && curr->next) {
            if (curr->val == curr->next->val) {
                int duplicateValue = curr->val;  // 找到重复的值，和重复的值相同的就删掉结点
                while (curr && curr->val == duplicateValue) {
                    ListNode* temp = curr;
                    curr = curr->next;
                    delete temp; // 释放内存
                }
                prev->next = curr;
            } else {
                prev = curr;
                curr = curr->next;
            }
        }

        return dummy.next;
    }
};