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
*/