题解 | #从单向链表中删除指定值的节点#

#include<iostream>
#include<vector>
using namespace std;

struct ListNode{
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr){}
};

ListNode* Delete_node(ListNode* head, int val){
    ListNode* dummyhead = new ListNode(0);
    dummyhead->next = head;
    ListNode* cur = dummyhead;
    while(cur!=nullptr){
        if(cur->next->val==val){
            cur->next=cur->next->next;
            return head;
        }
        cur=cur->next;
    }
    return head;
}

int main() {
    int num;//节点数
    cin>>num;
    int head; //头节点值
    cin>> head;
    ListNode* headnode = new ListNode(head);
    int follow;
    int front;
    vector<ListNode*> nodes;
    nodes.push_back(headnode);
    for(int i=0; i<2*(num-1); i+=2){
        cin>>follow>>front;
        ListNode* node_next = new ListNode(follow);
        int j=0;
        while(nodes[j]->val != front && j<nodes.size()){
            j++;
        }
        ListNode* buff = nodes[j]->next;
        nodes[j]->next= node_next;
        node_next->next = buff;
        nodes.push_back(node_next);
    }
    int del_val;
    cin>> del_val;
    headnode = Delete_node(headnode, del_val);
    //output
    //
    ListNode* cur=headnode;
    while(cur!=nullptr){
        cout<<cur->val<<" ";
        cur=cur->next;
    }
    return 0;
}
// 64 位输出请用 printf("%lld")