在单链表中删除指定值的节点

#include "List.h"
#include <stack>
using namespace std;
//使用栈，时间复杂度N，空间复杂度N
Node* removeN1(Node* head, int num)
{
    stack<Node*> nStack;
    while(head)
    {
        if(head->value != num)
        nStack.push(head);
        head = head->next;
    }

    while(!nStack.empty())
    {
        Node* cur = nStack.top();
        cur->next = head;
        head = cur;
        nStack.pop();
    }
    return head;
}
Node* removeN2(Node* head, int num)
{
    while(head)
    {
        if(head->value != num)
            break;
        head = head->next;
    }
    Node* pre = head;
    Node* cur = head;
    while(cur)
    {
        if(cur->value == num)
            pre->next = cur->next;
        else
            pre = cur;
        cur = cur->next;
    }
    return head;
}

int main()
{
    Node* pNode0 = new Node(0);
    Node* pNode1 = new Node(1, pNode0);
    Node* pNode2 = new Node(2, pNode1);
    Node* pNode3 = new Node(2, pNode2);
    Node* pNode4 = new Node(1, pNode3);
    Node* pNode5 = new Node(0, pNode4);

    pNode5 = removeN1(pNode5, 1);
    Print(pNode5);
    cout << "===============" << endl;
    pNode5 = removeN2(pNode5, 0);
    Print(pNode5);
}