题解 | #从单向链表中删除指定值的节点#

1. 记住列表创建得套路。
2. 记住cur得用法（每次要从头开始搜索）
3. 删列表中元素得时候，cur要指向dummy
4. 记住cur 以及 cur->next用来判断时，要注意cur->next为空，且要放到左边

#include<bits/stdc++.h>

using namespace std;
struct ListNode{
    int val;
    ListNode* next;

    ListNode(int x): val(x), next(NULL){

    }

};

int main(){
    int n,a,b;
    while(cin>>n){
        cin>>a;
        ListNode* dummy_head = new ListNode(-1);//头节点
        dummy_head->next = new ListNode(a);
        ListNode* cur =  dummy_head->next;

        //继续完成剩余节点得构建
        for(int i = 1; i< n;i++){
            cin>>a;
            cin>>b;
            cur = dummy_head->next;//每次要从第一个开始找
            while(cur->val!= b) cur = cur->next;

            ListNode* q = cur->next;//

            ListNode* p = new ListNode(a);
            cur->next = p;
            p->next = q;

        }



        int del;

        cin>>del;
        //删除

        ListNode* cur2 = dummy_head;

        //del
        while(cur2){

            if(cur2->next!=NULL && cur2->next->val==del ){//注意这种前瞻性判断， 且放到左边！！！
                cur2->next = cur2->next->next;
            }else{
                cur2 = cur2->next;
            }

        }


        while(dummy_head->next){
            if(dummy_head->next->next==NULL){
                cout<<dummy_head->next->val<<endl;
            }else{
                cout<<dummy_head->next->val<<" ";
            }

            dummy_head->next = dummy_head->next->next;
        }




    }

    return 0;
}