数据结构单链表删除操作解决方案

数据结构单链表删除操作

C/C++ code

#include <iostream>using namespace std;struct Node{    int num;    Node* pNext;};Node* pHead = NULL;Node* pEnd = NULL;void CreatList(int n)//创建链表{    Node* temp = new Node;    temp->num = n;    temp->pNext = NULL;    if(!pHead)    {        pHead = temp;    }    else    {        pEnd->pNext = temp;    }    pEnd = temp;}void DelList(int n){    Node* temp = pHead;    while(temp)//当节点不为空    {        if(temp!=pEnd)//如果不是尾结点，temp = temp->pNext        {            if(temp->num == n)//找到则删除            {                if(temp == pHead)//如果删除的是头结点                {                    pHead = temp->pNext;                    delete temp;                }                else             //如果删除的是中间节点                {                    Node* TempLast;                    Node* LinShi;                    TempLast->pNext = temp;                    LinShi = temp->pNext;                    TempLast->pNext = LinShi;                    delete temp;                }            }            else//如果删除的是尾节点            {                Node* LinShi;                LinShi->pNext = temp;                pEnd = LinShi;                delete temp;            }        }        temp = temp->pNext;    }}void ShowList(){    Node* temp = pHead;    while(temp)    {        cout<<temp->num<<" ";        temp = temp->pNext;    }    cout<<endl;}int main(){    CreatList(1);    CreatList(2);    CreatList(3);    CreatList(4);    CreatList(5);    CreatList(6);    ShowList();    DelList(1);    ShowList();    system("pause");    return 0;}

C/C++ code

temp = temp->pNext;//你这儿错了，上边都把temp删掉了，这儿肯定出错啊
[解决办法]
晕 上面代码格式没有弄好 再发一个
C/C++ code
struct Node{    int num;    Node* pNext;};Node* pHead = NULL;Node* pEnd = NULL;void CreatList(int n)//创建链表{    Node* temp = new Node;    temp->num = n;    temp->pNext = NULL;    if(!pHead)    {        pHead = temp;    }    else    {        pEnd->pNext = temp;    }    pEnd = temp;}void DelList(int n){    Node* temp = pHead;  // 指定需要删除的节点    Node* tempBefor=NULL; // 指定需要删除的节点的前一个    while(temp)//当节点不为空    {        if(temp!=pEnd)//如果不是尾结点，temp = temp->pNext        {            if(temp->num == n)//找到则删除            {                if(temp == pHead)//如果删除的是头结点                {                    pHead = temp->pNext;                    delete temp;                    break;                }                else             //如果删除的是中间节点                {                    tempBefor->pNext=temp->pNext;                                       delete temp;                    break;                }            }            else//如果删除的是尾节点            {                Node* LinShi;                LinShi->pNext = temp;                pEnd = LinShi;                delete temp;                break;            }                    }        //没找到则继续        tempBefor=temp;        temp = temp->pNext;    }}void ShowList(){    Node* temp = pHead;    while(temp)    {        cout<<temp->num<<" ";        temp = temp->pNext;    }    cout<<endl;}
[解决办法]
删除单链表的其中某一个时，如果不是头结点，那么就要先搞个节点指向他的前一个，
1     ->     2     -> 3
pre         temp     
如上如果查到的是要删除temp，那么只需要pre->next = temp->next;然后delete temp;就好了