leetcode——82. Remove Duplicates from Sorted List II

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该博客讨论了如何删除已排序链表中所有重复的节点，只保留原始列表中的唯一数字。提供了一个C++解决方案，该解决方案首先创建一个新的虚拟头节点，然后遍历链表，删除所有连续重复的节点，最后返回新链表的头节点。示例输入和输出展示了不同情况下的链表处理结果。

描述

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.
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Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]

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Input: head = [1,1,1,2,3]
Output: [2,3]

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head == nullptr || head->next == nullptr)
            return head;
        ListNode ans(0);
        ListNode *first = &ans;
        ans.next = head;
        ListNode *prev = first->next;
        ListNode *cur = prev->next;
        if(prev->val == cur->val && cur->next == nullptr)
            return nullptr;
        while(cur){
            bool del = false;
            while(cur->val == prev->val && cur->next !=nullptr){
                prev = cur;
                cur = cur->next;
                del = true;
            }
            if(prev->val == cur->val && cur->next == nullptr){
                // delete cur;
                // delete prev;
                first->next = nullptr;
                return ans.next;
            }
            if(del){
                first->next = cur;
                prev = cur;
                cur = cur->next;
            }
            else {
                first = first->next;
                prev = cur;
                cur = cur->next;
            }
        }
        return ans.next;
    }
};