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) {
        // write code here
        if(head == NULL)
            return NULL;
        ListNode* res = new ListNode(0);
        //在链表前加一个表头
        res->next = head;
        ListNode* cur = res;
         while(cur->next != NULL && cur->next->next != NULL){
            //遇到相邻两个节点值相同
            if(cur->next->val == cur->next->next->val){
                int temp = cur->next->val;
                //将所有相同的都跳过
                while (cur->next != NULL && cur->next->val == temp)
                    cur->next = cur->next->next;
            }
            else
                cur = cur->next;
        }
        //返回时去掉表头
        return res->next;
    }
};