O(1)时间内删除指定链表结点

题目

给定单链表头指针和一个结点指针，定义一个函数在O(1)时间内删除该结点。

分析

对于上图实例链表（a）删除指针p有两种方式

• 思路1：(b)找到前一个指针pre，赋值pre->next = p->next，删掉p
• 思路2：(c)目的是删除p，但是不删p，直接用p->next的值赋值给p，把p->next删除掉（好处：不用遍历找到p的前一个指针pre,O（1）时间内搞定）

于是，定位到思路2，但是思路2有两个特例：

1. 删除的是尾指针，需要遍历找到前一个指针
2. 整个链表就一个结点（属于删尾指针，但没法找到前面的指针，需要开小灶单独处理）

大体算法思路

待删指针不是尾指针：
      待删指针的值用待删指针的下一个指针的值覆盖
      删掉待删指针的下一个指针
待删指针是尾指针：
      待删指针同时是头指针：
            删掉头指针
     待删指针不是头指针
            找到待删指针的前一个指针
            删掉待删指针，前一个指针的next赋值为空

参考代码

bool deleteNode(ListNode *&head, ListNode *p)
{
    if(!p || !head)
        return false;
    if(p->m_pNext != NULL)   //不是尾指针
    {
        ListNode *del = p->m_pNext;
        p->m_nValue = del->m_nValue;
        p->m_pNext = del->m_pNext;
        delete del;
        del = NULL;
    }
    else if(head == p)       //是尾指针，同时只有一个结点
    {
        delete p;
        head = NULL;
    }
    else                     //是尾指针，同时有多个结点
    {
        ListNode *tmp = NULL, *pre = head;
        while(pre->m_pNext != p)
        {
            pre = pre->m_pNext;
        }
        delete p;
        p = NULL;
        pre->m_pNext = NULL;
    }
    return true;
}

完整代码1

#include <iostream>
using namespace std;

struct ListNode
{
    int m_nValue;
    ListNode* m_pNext;
};

bool deleteNode(ListNode *&head, ListNode *p)
{
    if(!p || !head)
        return false;
    if(p->m_pNext != NULL)
    {
        ListNode *del = p->m_pNext;
        p->m_nValue = del->m_nValue;
        p->m_pNext = del->m_pNext;
        delete del;
        del = NULL;
    }
    else if(head == p)
    {
        delete p;
        head = NULL;
    }
    else
    {
        ListNode *tmp = NULL, *pre = head;
        while(pre->m_pNext != p)
        {
            pre = pre->m_pNext;
        }
        delete p;
        p = NULL;
        pre->m_pNext = NULL;
    }
    return true;
}
void createList(ListNode *&head)
{
    head = new(ListNode);
    head->m_nValue = 1;
    head->m_pNext = NULL;

    ListNode *p2 = new(ListNode);
    p2->m_nValue = 2;
    p2->m_pNext = NULL;
    head->m_pNext = p2;

    ListNode *p3 = new(ListNode);
    p3->m_nValue = 3;
    p3->m_pNext = NULL;
    p2->m_pNext = p3;

    ListNode *p4 = new(ListNode);
    p4->m_nValue = 4;
    p4->m_pNext = NULL;
    p3->m_pNext = p4;
}

void deleteList(ListNode *p)
{
    ListNode *next = NULL;
    while(p != NULL)
    {
        cout << p->m_nValue << endl;
        next = p->m_pNext;
        delete p;
        p = NULL;
        p = next;
    }
}

int main()
{
    ListNode *head = NULL;
    createList(head);
    ListNode *p = head->m_pNext->m_pNext->m_pNext;
    deleteNode(head, p);
    deleteList(head);
}
    

完整代码2

#include <iostream>
using namespace std;

struct ListNode
{
    int m_nValue;
    ListNode* m_pNext;
};

bool deleteNode(ListNode **head, ListNode *p)
{
    if(!p || !head)
        return false;
    if(p->m_pNext != NULL)
    {
        ListNode *del = p->m_pNext;
        p->m_nValue = del->m_nValue;
        p->m_pNext = del->m_pNext;
        delete del;
        del = NULL;
    }
    else if(*head == p)
    {
        delete p;
        *head = NULL;
    }
    else
    {
        ListNode *tmp = NULL, *pre = *head;
        while(pre->m_pNext != p)
        {
            pre = pre->m_pNext;
        }
        delete p;
        p = NULL;
        pre->m_pNext = NULL;
    }
    return true;
}
void createList(ListNode *&head)
{
    head = new(ListNode);
    head->m_nValue = 1;
    head->m_pNext = NULL;

    ListNode *p2 = new(ListNode);
    p2->m_nValue = 2;
    p2->m_pNext = NULL;
    head->m_pNext = p2;

    ListNode *p3 = new(ListNode);
    p3->m_nValue = 3;
    p3->m_pNext = NULL;
    p2->m_pNext = p3;

    ListNode *p4 = new(ListNode);
    p4->m_nValue = 4;
    p4->m_pNext = NULL;
    p3->m_pNext = p4;
}

void deleteList(ListNode *p)
{
    ListNode *next = NULL;
    while(p != NULL)
    {
        cout << p->m_nValue << endl;
        next = p->m_pNext;
        delete p;
        p = NULL;
        p = next;
    }
}

int main()
{
    ListNode *head = NULL;
    createList(head);
    ListNode *p = head->m_pNext->m_pNext;
    deleteNode(&head, p);
    deleteList(head);
}
    

分析

删除非尾结点时间复杂读为O(1)，删除尾结点的时间复杂读为O(n), 平均时间复杂度为[（n-1）*O（1） + O(N)] / N = O(1)

还有删除函数并不能处理待删结点就是该链表中的指针，因此需要人为调用时控制，如果得验证的话，那就没必要做这些处理了，直接O(N)

技术细节——传值操作

#include <iostream>
using namespace std;

struct ListNode
{
    int m_nValue;
    ListNode* m_pNext;
};

void createList(ListNode *head)
{
    head = new(ListNode);
    head->m_nValue = 1;
    head->m_pNext = NULL;
}
void deleteList(ListNode *p)
{
    ListNode *next = NULL;
    while(p != NULL)
    {
        cout << p->m_nValue << endl;
        next = p->m_pNext;
        delete p;
        p = NULL;
        p = next;
    }
}

int main()
{
    ListNode *head = NULL;
    createList(head);
    cout << head << endl;
    deleteList(head);
}
    

主函数中的指针head为传值调用，传到函数并没有改变主函数中的值，图示

改进的措施就是引用传值，直接操纵原指针。