剑指 Offer 18. 删除链表的节点

参考书上解法：
问题：链表 *** 作不熟悉，需要强加练习！

class Solution {
public:
    ListNode* deleteNode(ListNode* head, int val) {
        ListNode* pnode = nullptr;
        ListNode* pre = new ListNode;
        if(!head) return pnode; // 如果是空链表直接返回空链表
        //链表只有一个节点，删除头节点也是尾节点
        if(head->val==val&&head->next==nullptr){
            ListNode* temp = new ListNode;
            temp = head;
            head = nullptr;
            // delete temp;
            return head;
        }
        pnode=head;
        while(pnode->next!=nullptr){
            //要删除的节点不是尾节点
            pre = pnode;
            ListNode* previous = new ListNode;
            if(pnode->val==val&&pnode->next!=nullptr){
                ListNode* current = new ListNode;
                ListNode* pnext = new ListNode;
                current = pnode;
                pnext = current->next;
                current->val = pnext->val;
                current->next = pnext->next;
                break;
            }
            pnode=pnode->next;
        }
        //要删除的节点是尾节点
        ListNode* cur = new ListNode;
        if(pnode->val==val){
            // ListNode* cur = pnode;
            // tail = previous;
            cur = pre;
            pre->next = nullptr;
        }
        return head;
    }
};

参考资料：
C++ single list
C single list