题解 | #删除链表的节点#

/**
 * 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) {
        // write code here
        if(head == nullptr)
            return nullptr;
        
        ListNode *res = new ListNode(0);
        res->next = head;

        ListNode *cur = res;

        while (cur->next!=nullptr && cur->next->val != val) {
            cur = cur->next;
        }
        if (cur->next->val == val) {
            cur->next = cur->next->next;
        }
        return res->next;
    }
};