题解 | #删除链表的节点#
删除链表的节点
https://www.nowcoder.com/practice/f9f78ca89ad643c99701a7142bd59f5d
/** * 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 */ //一个指针就可以 ListNode* deleteNode(ListNode* head, int val) { if (head == nullptr){ //边界处理 return head; } // write code here ListNode* dummy = new ListNode(0); //虚拟头节点,方便处理首个节点 dummy -> next = head; ListNode* cur = dummy; while (cur -> next != nullptr){ //终止条件 if (cur -> next -> val == val){ //删除节点,题目要求不考虑内存释放 cur -> next = cur -> next -> next; break; } cur = cur -> next; } return dummy -> next; } };