题解 | #删除有序链表中重复的元素-II#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* deleteDuplicates(ListNode* head) {
        // write code here
        if (!head) return nullptr;
        auto dummy = new ListNode(-1), cur = dummy;
        dummy->next = head;
        while (cur->next && cur->next->next) {
            if (cur->next->val == cur->next->next->val) {
                auto temp = cur->next->val;
                while (cur->next && cur->next->val == temp)
                    cur->next = cur->next->next;
            } else {
                cur = cur->next;
            }
        }
        return dummy->next;
    }
};

思路：与上一题的区别是删除所有出现次数大于1的节点，上一题保留一个出现次数大于1的节点

遍历cur的next节点，当cur的next节点值与cur的next的next节点值相等时，删除所有与cur的next节点值相同的节点

时间复杂度：O(n)

空间复杂度：O(1)