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 == nullptr) return head;

        map<int, int> hash_table;

        ListNode* pnode0 = head;
        ListNode* res = nullptr;

        while (pnode0 != nullptr) {
            int key = pnode0->val;
            if (hash_table.find(key) == hash_table.end()) {
                hash_table[key] = 1;
            } else {
                int value = hash_table[key] + 1;
                hash_table[key] = value;
            }
            pnode0 = pnode0->next;
        }

        for (auto it = hash_table.begin(); it != hash_table.end(); it++) {
            if (it->second == 1) {
                ListNode* newnode = new ListNode(it->first);
                if(res == nullptr) res = newnode;
                else{
                    ListNode* pnode = res;
                    while(pnode->next != nullptr) pnode = pnode->next;
                    pnode->next = newnode;
                }
            } else {
                continue;
            }
        }
        return res;
    }
};

map存储值和次数；

尾插法；构建新链表；