题解 | #删除链表的节点#
删除链表的节点
https://www.nowcoder.com/practice/f9f78ca89ad643c99701a7142bd59f5d
解法:创建链表虚拟头节点很重要
/** * 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 * @param val int整型 * @return ListNode类 */ ListNode* deleteNode(ListNode* head, int val) { if (head == nullptr) return nullptr; ListNode* dummyHead = new ListNode(0); dummyHead->next = head; ListNode* pCur = dummyHead; while (pCur->next) { if (pCur->next->val == val) { pCur->next = pCur->next->next; } else { pCur = pCur->next; } } ListNode* res = dummyHead->next; delete dummyHead; return res; } };
2023 剑指-链表 文章被收录于专栏
2023 剑指-链表