题解 | #牛牛的递增之旅#
牛牛的递增之旅
https://www.nowcoder.com/practice/e463addab7d548819d6b6483335651b5
/** * 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* removeDuplicates(ListNode* head) { // write code here if(nullptr == head) { return head; } ListNode* one_node = head; ListNode* two_node = head->next; while(nullptr != two_node) { if(one_node->val == two_node->val) { //更新后指针的位置 two_node = two_node->next; one_node->next = two_node; }else { one_node = two_node; two_node = two_node->next; } } return head; } };