链表的面试题

1、比较顺序表和链表的优缺点，它们分别在什么场景下使用？
1）顺序表支持随机访问，单链表不支持随机访问。
2）顺序表插入/删除数据效率很低，时间复杂度为O(N)（除尾插和尾删），单链表插入/删除效率更高，时间复杂度为O(1)。
3）顺序表的CPU高速缓冲效率更高，单链表CPU高速缓冲效率低。

2、打印单向链表

void Display(pList plist){    pNode cur = plist;    while (cur)    {         printf("%d-->", cur->data);         cur = cur->next;    }    printf("over\n");}

3、单链表的排序

void BubbleSort(pList *pplist){    pNode cur = NULL;    pNode tail = NULL;    assert(pplist);    if ((*pplist == NULL) || (*pplist)->next == NULL)    {        return;    }    cur = *pplist;    while (cur != tail)    {        while (cur->next != tail)        {            if (cur->data > cur->next->data)            {                DataType tmp = cur->data;                cur->data = cur->next->data;                cur->next->data = tmp;            }            cur = cur->next;        }        tail = cur;        cur = *pplist;    }}

4、逆序打印单项链表

void ReversePrint(pList plist){    pNode cur = plist;    if (cur == NULL)    {        return;    }    if (cur->next != NULL)    {        ReversePrint(cur->next);    }    printf("%3d", cur->data);}

5、删除无头单链表的非尾结点

void EraseNotTail(pNode pos){    pNode del = NULL;    assert(pos);    assert(pos->next != NULL);    pos->data = pos->next->data;    del = pos->next;    pos->next = pos->next->next;    free(del);    del = NULL;}

6、在无头单链表的非头结点前插入一个元素

void InsertFrontNode(pNode pos, DataType x){    pNode newnode = NULL;    DataType tmp;    assert(pos);    newnode = BuyNode(x);    newnode->next = pos->next;    pos->next = newnode;    tmp = pos->data;    pos->data = pos->next->data;    pos->next->data = tmp;} 

7、约瑟夫环问题

void JosephCycle(pList *pplist, int k){    pNode cur = NULL;    pNode del = NULL;    assert(pplist);    cur = *pplist;    while (cur->next != cur)//留一个数    {        int i = 0;        for (i = 0; i < k - 1; i++)        {            cur = cur->next;        }        del = cur->next;        printf("%d", cur->data);        cur->data = cur->next->data;        cur->next = cur->next->next;        free(del);        del = NULL;    }    printf("%d\n", cur->data);}

8、逆序单向链表

void ReverseList(pList *pplist){    DataType tmp = NULL;    pNode Next = NULL;    pNode cur = NULL;    assert(pplist);    if ((*pplist == NULL) || (*pplist)->next == NULL)    {        return;    }    cur = *pplist;    Next = cur->next;    cur->next = NULL;    while (Next != NULL)    {        tmp = Next->next;        Next->next = cur;        cur = Next;        Next = tmp;    }    *pplist = cur;}

9、合并两个有序列表

pList Merge(const pList *p1, const pList *p2){    pList newlist = NULL;    pList list1 = NULL;    pList list2 = NULL;    pList tail = NULL;    assert(p1 && p2);    list1 = *p1;    list2 = *p2;    if (list1 == list2)//同为一条链表，或都为空链表    {        return NULL;    }    if (list1 == NULL)    {        return list2;    }    if (list2 == NULL)    {        return list1;    }    if (list1->data < list2->data)//新链表的首部    {        newlist = list1;        list1 = list1->next;    }    else    {        newlist = list1;        list1 = list1->next;    }    tail = newlist;    while (list1 && list2)    {        if (list1->data && list2->data)        {            tail->next = list1;            list1 = list1->next;            tail = tail->next;        }        else        {            tail->next = list2;            list2 = list2->next;            tail = tail->next;        }    }    if (list1 = NULL)    {        tail->next = list2;    }    else    {        tail->next = list1;    }    return newlist;}

10、查找单链表的中间节点，要求只能遍历一次链表
//定义两个指针分别为fast和slow，fast每次跳两下，slow每次跳一下，
//当fast指向链尾时，show所指的位置就是中间结点

pNode FindMidNode(pList plist){    pNode fast = plist;    pNode slow = plist;    while (fast && (fast->next))    {        fast = fast->next->next;        slow = slow->next;    }    return slow;}

11、查找单链表的倒数第k个节点，要求只能遍历一次链表
//定义两个指针分别为fast和slow，让fast先走k-1步后slow再走，当fast指向尾部时，slow所指向的就是倒数第k个元素

pNode FindKNode(pList plist, int k){    pNode fast = plist;    pNode slow = plist;    while (fast && fast->next)    {        if (--k <= 0)//k次后，slow再走        {            slow = slow->next;        }        fast = fast->next;    }    if (k <= 0)//执行完while后才能输出    {        return slow;    }    return NULL;}

12、判断链表带环情况
//定义两个指针分别为fast和slow，每次fast比slow多走一步，如果带环，那么它们一定相交

pNode CheckCircle(pList plist){    pNode fast = plist;    pNode slow = plist;    while (fast && fast->next)    {        fast = fast->next->next;        slow = slow->next;        if (fast == slow)//带环        {            return fast;        }    }    return NULL;}

13、求环的长度

int GetCircleLength(pNode meet){    pNode pnode = meet->next;//meet为相遇点    int i = 1;    while (pnode != meet)    {        pnode = pnode->next;        i++;    }    return i;}

14、求环的入口点

pNode GetCycleEntryNode(pList plist, pNode meet){    pNode pnode = plist;    if ((plist == NULL) && (meet == NULL))    {        return NULL;    }    while(pnode != meet)    {        meet = meet->next;        pnode = pnode->next;    }    return meet;}

15、判断两条单项链表是否相交

int CheckCross(pList list1, pList list2){    pNode p1 = list1;    pNode p2 = list2;    while (p1 && p1->next)    {        p1 = p1->next;    }    while (p2 && p2->next)    {        p2 = p2->next;    }    if ((p1 == p2) && (p1 != NULL))//相交    {        return 1;    }    else        return -1;}

16、转化成不带环的链表求解

pNode GetCrossNoCycle(pList list1, pList list2){    pNode meet = NULL;    pNode cur = list1;    pNode crossNode = NULL;    while (cur && cur->next)    {        cur = cur->next;    }    cur->next = list2;    meet = CheckCircle(list1);    crossNode = GetCycleEnterNode(list1, meet);    return crossNode;}

17、求交点

pNode GetCrossNode(pList list1, pList list2){    pNode p1 = CheckCircle(list1);//1表带环；0表不带环    pNode p2 = CheckCircle(list2);    pNode crossNode = NULL;    if (p1 == NULL && p2 == NULL)//不带环    {        crossNode = GetCrossNoCycle(list1, list2);        return crossNode;    }    else if (p1 && p2)    {        pNode endNode = NULL;        pNode cur = NULL;        //找到环入口的第一个结点        endNode = GetCycleFrontEntryNode(list1, p1);        cur = endNode->next;//保存环的入口        endNode->next = NULL;//将环断开成两个不带环的链表        crossNode = GetCrossNoCycle(list1, list2);        endNode->next = cur;    }    return crossNode;}