AC算法每日打卡-删除链表的节点-正则表达式匹配

剑指 Offer 18. 删除链表的节点

class Solution {
public:
    ListNode* deleteNode(ListNode* head, int val) {
        ListNode* new_head = new ListNode(0);
        new_head->next = head;
        ListNode* cur = head;
        ListNode* ptr = new_head;
        while (cur != NULL && cur->val != val) {
            ptr = cur;
            cur = cur->next;
        }
        ptr->next = cur->next;
        return new_head->next;
    }
};

剑指 Offer 19. 正则表达式匹配

class Solution {
public:
    bool isMatch(string s, string p) {
        if(p.empty()) return s.empty();
        bool match=!s.empty()&&(s[0]==p[0] || p[0]=='.');
        if(p.length()>=2 && p[1]=='*'){
        
            return isMatch(s,p.substr(2)) || match && isMatch(s.substr(1),p);
        }else{
            if(match)
                return isMatch(s.substr(1),p.substr(1));
            else
                return false;
        }
    }
};