3.4 从无头单链表中删除节点

假设一个没有头指针的单链表，只有一个指向中间节点（不是头尾节点）的指针，请将该节点从单链表中删除。

思路：假设中间一段链表为 A->B->C，而p指向节点B。因为没有头节点，其实无法追溯回A，然后使A->C。所以这里的技巧是将C中的value值赋给B中，使B成为C，然后删除真正的C，从而达到目的。

#include "stdafx.h"
#include <iostream>

using namespace std;

const int MID_NUM = 3;
const int MAX_NUM = 8;

struct node{
    int value;
    struct node *next;
};

node* createMidLinkedList(int midNum,node *headPtr){
    node *midPtr,*p;

    for(int i=0;i<MAX_NUM;i++){
        p = (node *)malloc(sizeof(node));
        p->value = i;
        p->next = headPtr->next;
        headPtr->next = p;
        if(p->value==midNum)
            midPtr = p;
        cout<<p->value<<" -> ";
    }
    cout<<endl<<endl;
    return midPtr;
}

//delete middle node
bool deleteMidNode(node *midPtr){
    if(midPtr==NULL || midPtr->next==NULL) return false;

    node *mid,*midNext;
    mid = midPtr;
    midNext = mid->next;

    if(midNext!=NULL){
        mid->value = midNext->value;
        mid->next = midNext->next;
        delete midNext; //
    }
    

    return true;
}

int _tmain(int argc, _TCHAR* argv[])
{
    node *head= (node *)malloc(sizeof(node));; //不可在createMidLinkedList方法中分配内存
    head->next = NULL; //一定不可以漏，否则判断head->next==NULL时会返回false！
    node *mid = createMidLinkedList(MID_NUM,head);
    if(deleteMidNode(mid)){
        head = head->next;
        while(head!=NULL){
            cout<<head->value<<" -> ";

            head = head->next;
        }
    }else{
        cout<<"Failed to delete!"<<endl;
    }
}

输出为：

可见节点3(MID_NUM)已经被删除。

引申问题：只有一个头节点，翻转单链表，要求只遍历一次

利用构造链表时的方法，修改代码如下：

node *reverseLinkedList(node *headPtr){
    if(headPtr==NULL) return NULL;

    node *head,*newHead,*p,*temp;
    head=headPtr;
    p = head->next;
    newHead = (node *)malloc(sizeof(node));
    newHead->next = NULL;

    while(p!=NULL){
        temp = p->next;
        p->next    = newHead->next;
        newHead->next = p;
        p = temp;
    }

    return newHead;
}

int _tmain(int argc, _TCHAR* argv[])
{
    node *head= (node *)malloc(sizeof(node));; //不可在createMidLinkedList方法中分配内存
    head->next = NULL; //一定不可以漏，否则判断head->next==NULL时会返回false！
    node *mid = createMidLinkedList(MID_NUM,head);
    node *hPtr; //new header pointer
    if(deleteMidNode(mid)){
        hPtr = head;

        //print list
        head = head->next;
        while(head!=NULL){
            cout<<head->value<<" -> ";
            head = head->next;
        }
        cout<<endl<<endl;

        head = reverseLinkedList(hPtr);

        //print list
        head = head->next;
        while(head!=NULL){
            cout<<head->value<<" -> ";
            head = head->next;
        }

    }else{
        cout<<"Failed to delete!"<<endl;
    }
}

可见输出如下：