题解 | #从单向链表中删除指定值的节点#

#include <iostream>
#include <list>
using namespace std;
class llist{
    public:
        int num;
        llist* next;
        void add(llist &in){
            in.next=this->next;
            this->next=∈
        }
        void del(){
            if(this->next!=nullptr){
                llist *temp = this->next;
                this->next=this->next->next;
                delete temp;
            }
            else{
                delete this->next;
                this->next=nullptr;
            }
        }
        llist(int in):num(in),next(nullptr){}
};
int main() {
    int a;
    while (cin >> a) { // 注意 while 处理多个 case
        int first,b,c;
        cin>>first;
        llist *lt = new llist(first);
        for(int i=0;i<a-1;++i){
            llist* temp = lt;
            cin>>b>>c;
            while(temp!=nullptr){
                if(temp->num == c){
                    llist *temp1 = new llist(b);
                    temp->add(*temp1);
                    break;
                }
                temp=temp->next;
            } 
        }
        int del;
        cin>>del;
        llist* temp = lt;
        while (temp!=nullptr) {
            if(temp->next!=nullptr && temp->next->num == del){
                temp->del();
                break;
            }
            temp=temp->next;
        }
        while (lt!=nullptr){
            llist* temp = lt;
            cout<<lt->num<<' ';
            lt=lt->next;
            delete temp;
        }
    }
    return 0;
}
// 64 位输出请用 printf("%lld")