. - 力扣（LeetCode）

给定一个已排序的链表的头 head ， 删除原始链表中所有重复数字的节点，只留下不同的数字 。返回 已排序的链表 。

示例 1：

输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]

示例 2：

输入：head = [1,1,1,2,3]
输出：[2,3]

提示：

• 链表中节点数目在范围 [0, 300] 内
• -100 <= Node.val <= 100
• 题目数据保证链表已经按升序 排列

/**
 * 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 || !head->next) {
            return head;
        }
        ListNode* pre = nullptr;
        ListNode* cur = head;
        ListNode* returned_start = nullptr;
        ListNode* returned_end = nullptr;
        while (cur != nullptr) {
            if ((!pre || cur->val != pre->val) && (!cur->next || cur->val != cur->next->val)) {
                if (!returned_start) {
                    returned_start = cur;
                    returned_end = cur;
                } else {
                    returned_end->next = cur;
                    returned_end = cur;
                }
            }
            pre = cur;
            cur = cur->next;
        }
        returned_end->next = nullptr;
        return returned_start;
    }
};

class Solution {
   ListNode* deleteDuplicates(ListNode* head) {
        if(!head || !head->next) {
            return head;
        }
 
        ListNode* cur = new ListNode(0, head);
 
        while(cur->next && cur->next->next) {
            if(cur->next->val == cur->next->next->val) {
                ListNode* node = cur->next->next;
                while(node->next && node->val == node->next->val) {
                    node = node->next;
                }
                if(cur->next == head) {
                    head = node->next;
                }
                cur->next = node->next;
            } else {
                cur = cur->next;
            }
        }
 
        return head;
    }
}