终结Linked List（三）

第一篇终结Linked List（一）、终结Linked List（二）主要讲了单链表的基础知识，接下来的第二篇主要讲一些比较经典的问题。

一、Count()

给一个单链表和一个整数，返回这个整数在链表中出现了多少次。

/*
Given a list and an int, return the number of times
that int ocucurs in the list.
*/
int Count(struct node* head,int searchFor)
{
    int cnt = 0;
    struct node* cur = head;

    while (cur != NULL)
    {
        if (cur->data == searchFor)
            cnt++;
        cur = cur->next;
    }

    return cnt;
}

也可以用for循环实现。

二、GetNth()

给一个单链表和一个index，返回index位置上的数值，类似array[index]操作。

/*
Given a list and an index, return the data in the nth
node of the list. The nodes are numbered from 0.
Assert fails if the index is invalid (outside 0..length - 1).
*/
int GetNth(struct node* head,int index)
{
    int len = 0;
    struct node* cur = head;

    while (cur)
    {
        if (len == index)
        {
            return cur->data;
        }
        cur = cur->next;
        len++;
    }

    assert(0);  //如果走到这一行，表达式的值为假，断言失败
}

三、DeleteList()

给一个单链表，删除所有节点，使head为NULL。
删除链表{1,2,3}的示意图：

void DeleteList(struct node** headRef)
{
    struct node* cur = *headRef;  //deref headRef to get the real head

    while (*headRef)
    {
        cur = *headRef;
        *headRef = cur->next;
        free(cur);
    }
}

四、Pop()

给一个链表，删掉头节点，返回头节点的数据。
内存示意图：

/*
The opposite of Push().Takes a non-empty list and
remove the front node, and returns the data which was in that node.
*/
int pop(struct node** headRef)
{
    assert(*headRef != NULL);
    int ans = (*headRef)->data;  //pull out the data before the node is deleted

    struct node* cur = *headRef;
    *headRef = (*headRef)->next;   //unlink the head node for the caller
    free(cur);    //free the head node

    return ans;
}

五、InsertNth()

可以在[0,length]的任意位置插入指定元素。

/*
A more general version of Push().
Given a list, an index 'n' in the range 0..length,
and a data element, add a new node to the list so that
it has the given index.
*/
void InsertNth(struct node** headRef,int index,int data)
{
    //position 0 is a special case
    if (index == 0)
    {
        Push(headRef, data);
    }
    else
    {
        int cnt = 0;
        struct node* cur = *headRef;

        while (cnt < index - 1)
        {
            assert(cur != NULL);   //if this fails, the index was too big
            cur = cur->next;
            cnt++;
        }

        assert(cur != NULL);    //tricky:you have to check one last time

        Push(&(cur->next), data);
    }
}

这段代码坑有点多，可以通过画图或者单步跟踪的方法调试。
InsertNthTest()可以用来测试：

void InsertNthTest()
{
    struct node* head = NULL;   //start with the empty list

    InsertNth(&head, 0, 13);   //{13}
    InsertNth(&head, 1, 42);    //{13,42}
    InsertNth(&head, 1, 5);     //{13,5,42}
}

六、SortedInsert()

给定一个有序链表和一个节点，将该节点插入到合适的位置。
共有三种方法：
1、Uses special case code for the head end

void SortedInsert(struct node** headRef,struct node* newNode)
{
    //Special case for the head end
    if (newNode->data <= (*headRef)->data || *headRef == NULL)
    {
        newNode->next = *headRef;
        *headRef = newNode;
    }
    else
    {
        //Locate the node before the point of insertion
        struct node* cur = *headRef;
        while (cur->next && cur->next->data < newNode->data)
        {
            cur = cur->next;
        }
        newNode->next = cur->next;
        cur->next = newNode;
    }
}

2、Dummy node strategy for the head end
用dummy node这种方法一般不需要处理特殊情况。

void SortedInsert2(struct node** headRef,struct node* newNode)
{
    struct node dummy;
    struct node* cur = &dummy;
    dummy.next = *headRef;

    while (cur->next && newNode->data >= cur->next->data)
    {
        cur = cur->next;
    }
    newNode->next = cur->next;
    cur->next = newNode;

    *headRef = dummy.next;  //头指针永远指向dummy.next
}

3、Local references strategy for the head end

void SortedInsert3(struct node** headRef,struct node* newNode)
{
    struct node** curRef = headRef;

    while (*curRef && (*curRef)->data <= newNode->data)
    {
        curRef = &((*curRef)->next);
    }

    newNode->next = *curRef;  //Bug:(*curRef)->next  is incorrect

    *curRef = newNode;
}

原文地址：https://www.cnblogs.com/EIMadrigal/p/12234295.html