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) {
        if(head->val == val) return head->next;
        ListNode* pre = NULL, *cur = head;
        while(cur){
            if(cur->val == val){
                pre->next = cur->next;
            }
            pre = cur;
            cur = cur->next;
        }
        return head;
    }
};