题解 | #删除有序链表中重复的元素-I#
删除有序链表中重复的元素-I
https://www.nowcoder.com/practice/c087914fae584da886a0091e877f2c79
/** * 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) { // 空链表 if (head == nullptr) return nullptr; ListNode* dummy = new ListNode(-1); dummy->next = head; ListNode* pre = dummy; ListNode* cur = head; while (cur) { ListNode* next = cur->next; if (next == nullptr || cur->val != next->val) { cur->next = nullptr; pre->next = cur; pre = pre->next; } cur = next; } return dummy->next; } }; /* pre ↓ dummy(-1) → 1 → 1 → 2 → 2 → 3 → NULL ↑ ↑ cur next pre ↓ dummy(-1) → 1 → 1 → 2 → 2 → 3 → NULL ↑ ↑ cur next pre ↓--------------+ dummy(-1) → 1 → 1 (x) 2 → 2 → 3 → NULL | ↑ NULL next pre ↓ dummy(-1) → 1 → 1 (x) 2 → 2 → 3 → NULL ↑ ↑ cur next pre ↓ dummy(-1) → 1 → 1 (x) 2 → 2 → 3 → NULL ↑ ↑ cur next pre ↓-------------+ dummy(-1) → 1 → 1 (x) 2 → 2 (x) 3 → NULL | ↑ NULL next pre ↓ dummy(-1) → 1 → 1 (x) 2 → 2 (x) 3 → NULL ↑ ↑ cur next pre ↓-------+ dummy(-1) → 1 → 1 (x) 2 → 2 (x) 3 (x) NULL | ↑ NULL next pre ↓ dummy(-1) → 1 → 1 (x) 2 → 2 (x) 3 (x) NULL ↑ cur */